Class 12 - Maths - Continuity Differentiability

                                                                           Exercise 5.1

Question 1:

Prove that the function f(x) = 5x – 3 is continuous at x = 0, x = -3 and x = 5

Answer:

The given function is f(x) = 5x – 3

At x = 0, f(0) = 5 * 0 – 3 = -3

limx->0 f(x) = limx->0 (5x - 3) = 5 * 0 – 3 = -3

Therefore, f is continuous at x = 0

At x = -3, f(-3) = 5 * (-3) – 3 = -15 – 3 = -18

limx->-3 f(x) = limx->-3 (5x - 3) = 5 * (-3) – 3 = -15 – 3 = -18

Therefore, f is continuous at x = −3

At x = 5, f(5) = 5 * 5 – 3 = 25 – 3 = 22

limx->5 f(x) = limx->5 (5x - 3) = 5 * 5 – 3 = 25 – 3 = 22

Therefore, f(x) is continuous at x = 5.

Question 2:

Examine the continuity of the function f(x) = 2x2 – 1 at x = 3.

Answer:

The given function is f(x) = 2x2 – 1

At x = 3, f(3) = 2 * 32 – 1 = 2 * 9 – 1 = 18 – 1 = 17

limx->3 f(x) = limx->3 (2x2 – 1) = 2 * 32 – 1 = 2 * 9 – 1 = 18 – 1 = 17

Therefore, f(x) is continuous at x = 3.

Question 3:

Examine the following functions for continuity.

(a) f(x) = x – 5     (b) f(x) = 1/(x - 5), x ≠ 5      (c) f(x) = (x2 - 25)/(x + 5), x ≠ -5        (d) f(x) = |x - 5|

Answer:

(a) The given function is f(x) = x – 5

It is evident that f is defined at every real number k and its value at k is k − 5.

It is also observed that,

limx->k f(x) = limx->k (x - 5) = k – 5 = f(k)

So, limx->k f(x) = f(k)

Hence, f is continuous at every real number and therefore, it is a continuous function.

(b) The given function is f(x) = 1/(x - 5), x ≠ 5

For any real number k ≠ 5, we get

limx->k f(x) = limx->k 1/(x - 5) = 1/(k – 5)

Also, f(k) = 1/(k – 5)

So, limx->k f(x) = f(k)

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous

function.

(c) The given function is f(x) = (x2 - 25)/(x + 5), x ≠ -5

For any real number c ≠ −5, we get

limx->c f(x) = limx->c (x2 - 25)/(x + 5) = limx->c {(x - 5)(x + 5)}/(x + 5) = limx->c (x - 5) = c – 5

Also, f(c) = (c2 - 25)/(c + 5) = {(c - 5)(c + 5)}/(c + 5) = (c - 5)

So, limx->c f(x) = f(k)

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous

function.

(d) The given function is f(x) = |x - 5| =   (5 - x), if x < 5

                                                                        (x - 5), if x ≥ 5

This function f is defined at all points of the real line.

Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5

Case I: c < 5

Then, f(c) = 5 − c

limx->c f(x) = limx->c (5 - c) = 5 – c

So, limx->c f(x) = f(c)

Therefore, f is continuous at all real numbers less than 5.

Case II : c = 5

f(c) = f(5) = 5 – 5 = 0

limx->5- f(x) = limx->5 (5 - x) = 5 – 5 = 0

limx->5+ f(x) = limx->5 (x - 5) = 5 – 5 = 0

Therefore, f is continuous at x = 5

Case III: c > 5

Then, f (c) = 5 − c

limx->c f(x) = limx->c (x - 5) = c - 5

So, limx->c f(x) = f(c)

Therefore, f is continuous at all real numbers greater than 5.

Hence, f is continuous at every real number and therefore, it is a continuous function.

Question 4:

Prove that the function f(x) = xn is continuous at x = n, where n is a positive integer.

Answer:

The given function is f(x) = xn

It is evident that f is defined at all positive integers, n, and its value at n is nn

limx->n f(x) = limx->n xn = nn

So, limx->n f(x) = f(n)

Therefore, f is continuous at n, where n is a positive integer.

Question 5:

Is the function f defined by

f(x) =    x, if x ≤ 1

             5, if x > 1

continuous at x = 0? At x = 1? At x = 2?

Answer:

The given function f is

f(x) =    x, if x ≤ 1

             5, if x > 1

At x = 0,

It is evident that f is defined at 0 and its value at 0 is 0.

limx->0 f(x) = limx->0 x = 0

So, limx->0 f(x) = f(0)

Therefore, f is continuous at x = 0

At x = 1, f is defined at 1 and its value at 1 is 1.

The left hand limit of f at x = 1 is,

limx->1- f(x) = limx->1- x = 1

The right hand limit of f at x = 1 is,

limx->1+ f(x) = limx->1+ (5) = 5

So, limx->1- f(x) = limx->1+ f(x)

Therefore, f is not continuous at x = 1

At x = 2, f is defined at 2 and its value at 2 is 5.

limx->2 f(x) = limx->2 (5) = 5

So, limx->2 f(x) = f(2)

Therefore, f is continuous at x = 2

Question 6:

Find all points of discontinuity of f, where f is defined by

f(x) =    2x + 3, if x ≤ 2            

             2x - 3, if x > 2            

Answer:

The given function f is

f(x) =    2x + 3, if x ≤ 2             

             2x - 3, if x > 2               

It is evident that the given function f is defined at all the points of the real line.

Let c be a point on the real line. Then, three cases arise.

(i) c < 2                                         (ii) c > 2                              (iii) c = 2

Case (i) c < 2

Then, f(c) = 2c + 3

limx->c f(x) = limx->c (2x + 3) = 2c + 3

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 2

Case (ii) c > 2

Then, f(c) = 2c - 3

limx->c f(x) = limx->c (2x - 3) = 2c - 3

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 2

Case (iii) c = 2

Then, the left hand limit of f at x = 2 is,

limx->c- f(x) = limx->2- (2x + 3) = 2 * 2 + 3 = 4 + 3 = 7

The right hand limit of f at x = 2 is,

limx->c+ f(x) = limx->2+ (2x - 3) = 2 * 2 – 3 = 4 – 3 = 1

It is observed that the left and right hand limit of f at x = 2 do not coincide.

Therefore, f is not continuous at x = 2

Hence, x = 2 is the only point of discontinuity of f.

Question 7:

Find all points of discontinuity of f, where f is defined by

f(x) =  { |x| + 3, if x ≤ -3 ,

            -2x , if -3 < x < 3 ,

            6x + 2, if x ≥ 3 }

Answer:

The given function f is

 f(x) =   { |x| + 3, if x ≤ -3

             -2x , if -3 < x < 3

             6x + 2, if x ≥ 3 }

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If c < -3 then f(c) = -c + 3

limx->c f(x) = limx->c (-x + 3) = -c + 3

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < −3

Case II:

If c = -3 then f(-3) = -(-3) + 3 = 3 + 3 = 6

limx->-3- f(x) = limx->-3- (-x + 3) = -(-3) + 3 = 3 + 3 = 6

limx->-3+ f(x) = limx->-3+ (-2x) = -2 * (-3) = 6

Therefore, f is continuous at x = −3

Case III:

If -3 < c < 3, then

f(c) = -2c and limx->c f(x) = limx->c (-2x) = -2c

So, limx->c f(x) = f(c)

Therefore, f is continuous in (−3, 3).

Case IV:

If c = 3, then the left hand limit of f at x = 3 is,

limx->3- f(x) = limx->3- (-2x) = -2 * 3 =6

The right hand limit of f at x = 3 is,

limx->3+ f(x) = limx->3+ (6x + 2) = 6 * 3 + 2 = 18 + 2 = 20

It is observed that the left and right hand limit of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3

Case V:

If c > 3, then

f(c) = 6c + 2 and limx->c f(x) = limx->c (6x + 2) = 6c + 2

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 3

Hence, x = 3 is the only point of discontinuity of f.

Question 8:

Find all points of discontinuity of f, where f is defined by

f(x) =   { |x|/x, if x ≠ 0

             0, if x = 0 }

Answer:

The given function f is

f(x) =   { |x|/x, if x ≠ 0 ,

             0, if x = 0 }

It is known that,

        x < 0 => |x| = -x

and x > 0 => |x| = x

Therefore, the given function can be rewritten as

f(x) =    |x|/x = -x/x = -1, if x < 0 ,

              0, if x = 0 ,

              |x|/x = x/x = 1, if x > 0 }          

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If c < 0, then f(c) = -1

limx->c f(x) = limx->c (-1) = -1

Therefore, f is continuous at all points x < 0

Case II:

If c = 0, then the left hand limit of f at x = 0 is,

limx->0- f(x) = limx->0- (-1) = -1

The right hand limit of f at x = 0 is,

limx->0+ f(x) = limx->0+ (1) = 1

It is observed that the left and right hand limit of f at x = 0 do not coincide.

Therefore, f is not continuous at x = 0

Case III:

If c > 0, then f(c) = 1

limx->c f(x) = limx->c f(1) = 1

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 0

Hence, x = 0 is the only point of discontinuity of f.

Question 9:

Find all points of discontinuity of f, where f is defined by

f(x) =    {x/|x|, if x < 0 ,

             -1, if x ≥ 0 }

Answer:

The given function f is

f(x) =   { x/|x|, if x < 0 ,

             -1, if x ≥ 0 }

It is known that, if x < 0 => |x| = -x

Therefore, the given function can be rewritten as

f(x) =   { x/|x| = x/(-x) = -1, if x < 0 ,

             -1, if x ≥ 0 }

=> f(x) = -1 for all x Є R

Let c be any real number. Then,

limx->c f(x) = limx->c f(-1) = -1

Also, f(c) = -1 = limx->c f(x)

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Question 10:

Find all points of discontinuity of f, where f is defined by

f(x) =   { x + 1, if x ≥ 1 ,

             x2 + 1, if x < 1 }

 Answer:

The given function f is

f(x) =   { x + 1, if x ≥ 1 ,

             x2 + 1, if x < 1 }

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If c < 1, then f(c) = c2 + 1

And limx->c f(x) = limx->c (x2 + 1) = c2 + 1

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then f(c) = f(1) = 1 + 1 = 2

The left hand limit of f at x = 1 is,

limx->1- f(x) = limx->1- (x2 + 1) = 12 + 1 = 2

The right hand limit of f at x = 1 is,

limx->1+ f(x) = limx->1+ (x + 1) = 1 + 1 = 2

Therefore, f is continuous at x = 1

Case III:

If c > 1, then f(c) = c + 1

limx->c f(x) = limx->c (x + 1) = c + 1

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 1

Hence, the given function f has no point of discontinuity.

Question 11:

Find all points of discontinuity of f, where f is defined by

f(x) =   { x3 - 3, if x ≤ 2 ,

             x2 + 1, if x > 2 }

Answer:

The given function f is

f(x) =   { x3 - 3, if x ≤ 2

             x2 + 1, if x > 2 }

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If c < 2, then f(c) = c3 - 3

And limx->c f(x) = limx->c (x3 - 3) = c3 - 3

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 2

Case II:

If c = 2, then f(c) = f(2) = 23 - 3 = 8 – 3 = 5

The left hand limit of f at x = 2 is,

limx->2- f(x) = limx->2- (x3 - 3) = 23 - 3 = 8 – 3 = 5

The right hand limit of f at x = 1 is,

limx->2+ f(x) = limx->2+ (x2 + 1) = 22 + 1 = 4 + 1 = 5

Therefore, f is continuous at x = 2

Case III:

If c > 2, then f(c) = c2 + 1

limx->c f(x) = limx->c (x2 + 1) = c2 + 1

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 2

Thus, the given function f is continuous at every point on the real line.

Hence, f has no point of discontinuity.

Question 12:

Find all points of discontinuity of f, where f is defined by

f(x) =   { x10 - 1, if x ≤ 1 ,

             x2,        if x > 1 }

Answer:

The given function f is

f(x) =   { x10 - 1, if x ≤ 1 ,

             x2,        if x > 1 }

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If c < 1, then f(c) = c10 - 1

And limx->c f(x) = limx->c (x10 - 3) = c10 - 1

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

limx->1- f(x) = limx->1- (x10 - 1) = 110 - 1 = 1 – 1 = 0

The right hand limit of f at x = 1 is,

limx->1+ f(x) = limx->1+ (x2) = 12 = 1

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

If c > 1, then f(c) = c2

limx->c f(x) = limx->c (x2) = c2

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of

discontinuity of f.

Question 13:

Is the function defined by

f(x) =   { x + 5,  if x ≤ 1 ,

             x – 5,  if x > 1 }

a continuous function?

Answer:

The given function is

f(x) =   { x + 5,  if x ≤ 1 ,

             x – 5,  if x > 1 }

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If c < 1, then f(c) = c + 5

And limx->5 f(x) = limx->5 (x + 5) = c + 5

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then f(1) = 1 + 5 = 6

The left hand limit of f at x = 1 is,

limx->1- f(x) = limx->1- (x + 5) = 1 + 5 = 6

The right hand limit of f at x = 1 is,

limx->1+ f(x) = limx->1+ (x - 5) = 1 - 5 = -4

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

If c > 1, then f(c) = c - 5

limx->c f(x) = limx->c (x - 5) = c - 5

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of

discontinuity of f.

Question 14:

Discuss the continuity of the function f, where f is defined by

              { 3, if 0 ≤ x ≤ 1  ,

f(x) =      4, if 1 < x < 3 ,

               5, if 3 ≤ x ≤ 10 }

Answer:

The given function is

                3, if 0 ≤ x ≤ 1  

f(x) =      4, if 1 < x < 3

               5, if 3 ≤ x ≤ 10

The given function is defined at all points of the interval [0, 10].

Let c be a point in the interval [0, 10].

Case I:

If 0 ≤ c < 1, then f(c) = 3

And limx->c f(x) = limx->c (3) = 3

So, limx->c f(x) = f(c)

Therefore, f is continuous in the interval [0, 1).

Case II:

If c = 1, then f(3) = 3

The left hand limit of f at x = 1 is,

limx->1- f(x) = limx->1- (3) = 3

The right hand limit of f at x = 1 is,

limx->1+ f(x) = limx->1+ (4) = 4

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

If 1 < c < 3, then f(c) = 4

And limx->c f(x) = limx->c (4) = 4

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points of the interval (1, 3).

Case IV:

If c = 3, then f(c) = 5

The left hand limit of f at x = 3 is,

limx->3- f(x) = limx->3- (4) = 4

The right hand limit of f at x = 3 is,

limx->3+ f(x) = limx->3+ (5) = 5

It is observed that the left and right hand limits of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3

Case V:

If 3 < c < 10, then f(c) = 5

And limx->c f(x) = limx->c (5) = 5

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points of the interval (3, 10].

Hence, f is not continuous at x = 1 and x = 3

 Question 15:

Discuss the continuity of the function f, where f is defined by

            { 2x, if x < 0  ,

f(x) =      0, if 0 ≤ x ≤ 1,

               4x, if x > 1}

Answer:

Given function is

             { 2x, if x < 0  ,

f(x) =      0, if 0 ≤ x ≤ 1 ,

               4x, if x > 1 }

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

If c < 0, then f(c) = 2c

limx->c f(x) = limx->c (2x) = 2c

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 0

Case II:

If c = 0, the f(c) = f(0) = 9

The left hand limit of f at x = 0 is,

limx->0- f(x) = limx->0- (2x) = 2 * 0 = 0

The right hand limit of f at x = 0 is,

limx->0+ f(x) = limx->0+ (0) = 0

So, limx->0 f(x) = f(0)

Therefore, f is continuous at x = 0

Case III:

If 0 < c < 1, then f(0) = 0

And limx->c f(x) = limx->c (0) = 0

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points of the interval (0, 1).

Case IV:

If c = 1, the f(c) = f(1) = 0

The left hand limit of f at x = 1 is,

limx->1- f(x) = limx->1- (0) = 0

The right hand limit of f at x = 1 is,

limx->1+ f(x) = limx->1+ (4x) = 4 * 1 = 4

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case V:

If c < 1, then f(c) = 4c

And limx->c f(x) = limx->c (4x) = 4c

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 1

Hence, f is not continuous only at x = 1

Question 16:

Discuss the continuity of the function f, where f is defined by

             {  -2, if x ≤ -1  ,

f(x) =      2x, if -1 < x ≤ 1 ,

               2, if x > 1 }

Answer:

Given function is

              { -2, if x ≤ -1  ,

f(x) =      2x, if -1 < x ≤ 1 ,

               2, if x > 1 }

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

If c < -1, then f(c) = -2

limx->c f(x) = limx->c (-2) = -2

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < -1

Case II:

If c = -1, the f(c) = f(-1) = -2

The left hand limit of f at x = -1 is,

limx->-1- f(x) = limx->-1- (-2) = -2

The right hand limit of f at x = -1 is,

limx->-1+ f(x) = limx->-1+ (2x) = 2 * (-1) = -2

So, limx->0 f(x) = f(-1)

Therefore, f is continuous at x = -1

Case III:

If -1 < c < 1, then f(c) = 2c

And limx->c f(x) = limx->c (2x) = 2c

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points of the interval (-1, 1).

Case IV:

If c = 1, the f(c) = f(1) = 2 * 1 = 2

The left hand limit of f at x = 1 is,

limx->1- f(x) = limx->1- (2x) = 2 * 1 = 2

The right hand limit of f at x = 1 is,

limx->1+ f(x) = limx->1+ (2x) = 2 * 1 = 2

Since limx->1- f(x) = limx->1+ f(x) = 2

Therefore, f is continuous at x = 1

Case V:

If c > 1, then f(c) = 2

And limx->c f(x) = limx->c (2) = 2

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the

real line.

Question 17:

Find the relationship between a and b so that the function f defined by

f(x) =    {ax + 1, if x ≤ 3 ,

             bx + 3, if x > 3 }

is continuous at x = 3.

Answer:

The given function f is

f(x) =    {ax + 1, if x ≤ 3 ,

             bx + 3, if x > 3 }

If f is continuous at x = 3, then

      limx->3- f(x) = limx->3+ f(x) = 3a + 1   …………1

Also,

limx->3- f(x) = limx->3- (ax + 1) = 3a + 1

limx->3+ f(x) = limx->3+ (bx + 3) = 3b + 3

Therefore, from equation 1, we get

     3a + 1 = 3b + 3 = 3a + 1

=> 3a + 1 = 3b + 3

=> 3a = 3b + 2

=> a = 3b/3 + 2/3

=> a = b + 2/3

Therefore, the required relationship is given by, a = b + 2/3

 

Question 18:

For what value of λ is the function defined by

f(x) =    {λ(x2 – 2x), if x ≤ 0 ,

             4x + 1, if x > 0 }

is continuous at x = 0? What about continuity at x = 1?

Answer:

The given function f is

f(x) =   { λ(x2 – 2x), if x ≤ 0 ,

             4x + 1, if x > 0 }

If f is continuous at x = 0, then

     limx->0- f(x) = limx->0+ f(x) = f(0)

=> limx->0- {λ(x2 – 2x)} = limx->0+ (4x + 1) = λ(02 – 2 * 0)

=> λ * 0 = 4 * 0 + 1

=> 0 = 1, which is not possible.

Therefore, there is no value of λ for which f is continuous at x = 0

At x = 1,

f (1) = 4x + 1 = 4 * 1 + 1 = 5

limx->1 (4x + 1) = 4 * 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x = 1

Question 19:

Show that the function defined by g(x) = x – [x] is discontinuous at all integral point.

Here [x] denotes the greatest integer less than or equal to x.

Answer:

The given function is g(x) = x – [x]

It is evident that g is defined at all integral points.

Let n be an integer.

Then, g(n) = n – [n] = n – n = 0

The left hand limit of f at x = n is,

 limx->n- g(x) = limx->n- (x – [x]) = limx->n- (x) - limx->n- ([x]) = n – (n - 1) = n – n + 1 = 1

The right hand limit of f at x = n is,

limx->n+ g(x) = limx->n+ (x – [x]) = limx->n+ (x) - limx->n+ ([x]) = n – n = 0

It is observed that the left and right hand limits of f at x = n do not coincide.

Therefore, f is not continuous at x = n

Hence, g is discontinuous at all integral points.

Question 20:

Is the function defined by f(x) = x2 – sin x + 5 continuous at x = π?

Answer:

The given function is f(x) = x2 – sin x + 5

It is evident that f is defined at x = π

At x = π, f(x) = f(π) = π 2 – sin π + 5 = π 2 – 0 + 5 = π 2 + 5

Consider limx->π f(x) = limx->π (x2 – sin x + 5)

Put x = π + h

If x -> π, then it is evident that h->0

So, limx->π f(x) = limx->π (x2 – sin x + 5)

                         = limh->0 [(π + h)2 – sin (π + h) + 5]

                         = limh->0 (π + h)2 – limh->0 sin (π + h) + limh->0 5

                         = (π + 0)2 – limh->0 [sin π * cos h + cos π * sin h] + 5

                         = π2 – limh->0 [sin π * cos h] - limh->0 [cos π * sin h] + 5 

                         = π2 – sin π * cos 0 - cos π * sin 0 + 5

                         = π2 – 0 * 1 + (-1) * 0 + 5       

                          = π2 + 5

So, limx->π f(x) = f(π)

Therefore, the given function f is continuous at x = π

Question 21:

Discuss the continuity of the following functions.                                                                               

(a) f (x) = sin x + cos x                     (b) f (x) = sin x − cos x                           (c) f (x) = sin x * cos x

Answer:

It is known that if g and h are two continuous functions, then

g + h, g – h and g.h are also continuous.

It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.

Let g (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If x -> c, then h -> 0

g(c) = sin c

limx->c g(x) = limx->c sin x

                   = limh->0 sin (c + h)

                   = limh->0 [sin c * cos h + cos c * sin h]

                   = limh->0 [sin c * cos h] + limh->0 [cos c * sin h]

                   = sin c * cos 0 + cos c * sin 0

                   = sin c * 1 + cos c * 0

                   = sin c + 0

                   = sin c

So, limx->c g(x) = g(c)

Therefore, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x ->c, then h >0

h (c) = cos c

limx->c h(x) = limx->c cos x

                   = limh->0 cos (c + h)

                   = limh->0 [cos c * cos h - sin c * sin h]

                   = limh->0 [cos c * cos h] - limh->0 [sin c * sin h]

                   = cos c * cos 0 + sin c * sin 0

                   = cos c * 1 + sin c * 0

                   = cos c + 0

                   = cos c

So, limx->c h(x) = h(c)

Therefore, h is a continuous function.

Therefore, it can be concluded that

(a) f (x) = g (x) + h (x) = sin x + cos x is a continuous function

(b) f (x) = g (x) − h (x) = sin x − cos x is a continuous function

(c) f (x) = g (x) * h (x) = sin x * cos x is a continuous function

Question 22:

Discuss the continuity of the cosine, cosecant, secant and cotangent functions,

Answer:

It is known that if g and h are two continuous functions, then

(i) h(x)/g(x), g(x) ≠ 0 is continuous

(ii) 1/g(x), g(x) ≠ 0 is continuous

(iii) 1/h(x), h(x) ≠ 0 is continuous

It has to be proved first that g(x) = sin x and h(x) = cos x are continuous functions.

Let g(x) = sin x

It is evident that g(x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If x -> c then h -> 0

g(c) = sin c

limx->c g(x) = limx->c sin x

                   = limh->0 sin (c + h)

                   = limh->0 [sin c * cos h + cos c * sin h]

                   = limh->0 [sin c * cos h] + limh->0 [cos c * sin h]

                   = sin c * cos 0 + cos c * sin 0

                   = sin c * 1 + cos c * 0

                   = sin c + 0

                   = sin c

So, limx->c g(x) = g(c)

Therefore, g is a continuous function.

Let h(x) = cos x

It is evident that h(x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x  -> c, then h -> 0

h (c) = cos c

limx->c h(x) = limx->c cos x

                   = limh->0 cos (c + h)

                   = limh->0 [cos c * cos h - sin c * sin h]

                   = limh->0 [cos c * cos h] - limh->0 [sin c * sin h]

                   = cos c * cos 0 + sin c * sin 0

                   = cos c * 1 + sin c * 0

                   = cos c + 0

                   = cos c

So, limx->c h(x) = h(c)

Therefore, h (x) = cos x is continuous function.

It can be concluded that,

cosec x = 1/sin x, sin x ≠ 0 is continuous

=> cosec x, x ≠ nπ (n є Z) is continuous

Therefore, cosecant is continuous except at x = nπ, n є Z

sec x = 1/cos x, cos x ≠ 0 is continuous

=> sec x, x ≠ (2n + 1)π/2 (n є Z) is continuous

Therefore, secant is continuous except at x = (2n + 1)π/2 (n є Z)

cot x = sin x / cos x, sin x ≠ 0 is continuous

=> cot x, x ≠ nπ (n є Z) is continuous

Therefore, cotangent is continuous except at x = nπ (n є Z)

Question 23:

Find the points of discontinuity of f, where

f(x) =    sin x /x, if x < 0

              x + 1, if x ≥ 0

Answer:

The given function f is

f(x) =    sin x /x, if x < 0

              x + 1, if x ≥ 0

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

If c < 0, then f(c) = sin c /c

And limx->c f(x) = limx->c (sin x / x) = sin c / c

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 0

Case II:

If c > 0, then f(c) = c + 1

And limx->c f(x) = limx->c (x + 1) = c + 1

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 0

Case III:

If c = 0, then f(c) = f(0) = 0 + 1 = 1

The left hand limit of f at x = 0 is,

limx->0- f(x) = limx->0- (sin x / x) = 1

The right hand limit of f at x = 0 is,

limx->0+ f(x) = limx->0+ (x + 1) = 1

So, limx->0- f(x) = limx->0+ f(x) = f(0)

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at all points of the real

line. Thus, f has no point of discontinuity.

Question 24:

Determine if f defined by

f(x) =    x2 * sin 1/x, if x ≠ 0

             0,                  if x = 0

is a continuous function?

Answer:

The given function f is

f(x) =  {  x2 * sin 1/x, if x ≠ 0 ,

             0,                  if x = 0 }

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

If c ≠ 0, then f(c) = c2 * sin 1/c

 limx->c f(x) = limx->c (x2 * sin 1/x) = (limx->c x2) * (limx->c sin 1/x) = c2 * sin 1/c

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x ≠ 0

Case II:

If c = 0, the f(c) = 0

limx->0- f(x) = limx->0- (x2 * sin 1/x) = limx->0 (x2 * sin 1/x)

It is known that

      -1 ≤ sin 1/x ≤ 1, x ≠ 0

=> -x2 ≤ x2 * sin 1/x ≤ x2

=> limx->0 (-x2) ≤ limx->0 (x2 * sin 1/x) ≤ limx->0 (x2)

=> 0 ≤ limx->0 (x2 * sin 1/x) ≤ 0

=> limx->0 (x2 * sin 1/x) = 0

=> limx->0- f(x) = 0

Similarly, limx->0+ f(x) = limx->0+ (x2 * sin 1/x) = limx->0 (x2 * sin 1/x) = 0

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the

real line. Thus, f is a continuous function.

Question 25:

Examine the continuity of f, where f is defined by

f(x) =   { sin x – cos x, if x ≠ 0 ,

             -1, if x = 0 }

Answer:

The given function f is

 

f(x) =   { sin x – cos x, if x ≠ 0 ,

             -1, if x = 0 }

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

If c ≠ 0, then f(c) = sin c – cos c

 limx->c f(x) = limx->c (sin x – cos x) = sin c – cos c

So, limx->c f(x) = f(c)

Therefore, f is continuous at all points x, such that x ≠ 0

Case II:

If c = 0, then f(0) = sin 0 – cos 0 = -1

limx->0- f(x) = limx->0 (sin x – cos x) = sin 0 – cos 0 = -1

limx->0+ f(x) = limx->0 (sin x – cos x) = sin 0 – cos 0 = -1

So, limx->0- f(x) = limx->0+ f(x) = f(c)

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the

real line. Thus, f is a continuous function.

Question 26:

Find the values of k so that the function f is continuous at the indicated point.

f(x) =   { (k * cos x)/(π – 2x), if x ≠ π/2 ,

             3, if x = π/2 }   at x = π/2

Answer:

The given function f is

f(x) =   { (k * cos x)/(π – 2x), if x ≠ π/2 ,

             3, if x = π/2 }

The given function f is continuous at x = π/2, if f is defined at and if the value of the f at x = π/2

equals the limit of f at x = π/2

It is evident that f is defined at x = π/2 and f(π/2) = 3

limx->π/2 f(x) = limx->π/2 [(k * cos x)/(π – 2x)]

Put x = π/2 + h

Then x -> π/2

=> h -> 0

So, limx->π/2 f(x) = limx->π/2 [(k * cos x)/(π – 2x)]

                            = limh->0 [{k * cos (π/2 + h)}/{π – 2(π/2 + h)}]

                            = k * limh->0 [(-sin h)/(-2h)]

                            = (k/2) * limh->0 [sin h/h]

                            = (k/2) * 1

                            = k/2

So, limx->π/2 f(x) = f(π/2)   

=> k/2 = 3

=> k = 6

Therefore, the required value of k is 6.

Question 27:

Find the values of k so that the function f is continuous at the indicated point.

f(x) =   { kx2, if x ≤ 2 ,

             3, if x > 2 }                  at x = 2

Answer:

The given function is

f(x) =    kx2, if x ≤ 2

             3, if x > 2

The given function f is continuous at x = 2, if f is defined at x = 2 and if the value of f at x = 2

equals the limit of f at x = 2

It is evident that f is defind at x = 2 and f(2) = k(2)2 = k * 4 = 4k

      limx->2- f(x) = limx->2+ f(x) = f(2)

=> limx->2- (kx2) = limx->2+ (3) = 4k

=> k * 22 = 3 = 4k

=> 4k = 3

=> k = 3/4

Therefore, the required value of k is 3/4

Question 28:

Find the values of k so that the function f is continuous at the indicated point.

f(x) =   { kx + 1, if x ≤ π ,

             cos x, if x > π  }                                    at x = π

Answer:

The given function is

f(x) =    kx + 1, if x ≤ π

             cos x, if x > π

The given function f is continuous at x = π, if f is defined at x = π and if the value of f at x = π

equals the limit of f at x = π

It is evident that f is defind at x = π and f(π) = kπ + 1

      limx->π- f(x) = limx-> π+ f(x) = f(π)

=> limx->π- (kx + 1) = limx-> π+ cos x = kπ + 1

=> kπ + 1 = cos π = kπ + 1

=> kπ + 1 = -1 = kπ + 1

=> kπ + 1 = -1

=> kπ = -2

=> k = -2/π

Therefore, the required value of k is -2/π

Question 29:

Find the values of k so that the function f is continuous at the indicated point.

f(x) =  {  kx + 1, if x ≤ 5 ,

             3x - 5, if x > 5 }                                 at x = 5

Answer:

The given function is

f(x) =    kx + 1, if x ≤ 5

             3x - 5, if x > 5

The given function f is continuous at x = 5, if f is defined at x = 5 and if the value of f at x = 5

equals the limit of f at x = 5

It is evident that f is defind at x = 5 and f(5) = 5k + 1

      limx->5- f(x) = limx->5+ f(x) = f(5)

=> limx->5- (kx + 1) = limx->5+ (3x - 5) = 5k + 1

=> 5k + 1 = 15 - 5 = 5k + 1

=> 5k + 1 = 10

=> 5k = 9

=> k = 9/5

Therefore, the required value of k is 9/5

Question 30:

Find the values of a and b such that the function defined by

               { 5, if x ≤ 2  ,

f(x) =      ax + b, if 2 < x < 10 ,

               21, if x ≥ 10 }

is a continuous function.

Answer:

The given function f is

              { 5, if x ≤ 2  ,

f(x) =      ax + b, if 2 < x < 10 ,

               21, if x ≥ 10 }

It is evident that the given function f is defined at all points of the real line.

If f is a continuous function, then f is continuous at all real numbers.

In particular, f is continuous at x = 2 and x = 10

Since f is continuous at x = 2, we get

      limx->2- f(x) = limx->2+ f(x) = f(2)

=> limx->2- (5) = limx->2+ (ax + b) = 5

=> 5 = 2a + b = 5

=> 2a + b = 5     …………..1

Since f is continuous at x = 10, we get

      limx->10- f(x) = limx->10+ f(x) = f(10)

=> limx->10- (ax + b) = limx->10+ (21) = 21

=> 10a + b = 21 = 21

=> 10a + b = 21     …………..2

On subtracting equation 1 from equation 2, we get

     8a = 16

=> a = 2

By putting a = 2 in equation 1, we obtain

     2 * 2 + b = 5

=> 4 + b = 5

=> b = 1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.

Question 31:

Show that the function defined by f(x) = cos (x2) is a continuous function.

Answer:

The given function is f(x) = cos (x2)

f = g o h, where g (x) = cos x and h(x) = x2

[Since (goh)(x) = g(h(x)) = g(x2) = cos (x2) = f(x)]

It has to be first proved that g(x) = cos x and h(x) = x2 are continuous functions.

It is evident that g is defined for every real number.

Let c be a real number.

Then, g(c) = cos c

Put x = c + h

If x -> c, then h -> 0

limx->c g(x) = limx->c cos x

                   = limh->0 cos (c + h)

                   = limh->0 [cos c * cos h - sin c * sin h]

                   = limh->0 [cos c * cos h] - limh->0 [sin c * sin h]

                   = cos c * cos 0 + sin c * sin 0

                   = cos c * 1 + sin c * 0

                   = cos c + 0

                   = cos c

So, limx->c g(x) = g(c)

Therefore, g (x) = cos x is continuous function.

h(x) = x2

Clearly, h is defined for every real number.

Let k be a real number, then h(k) = k2

limx->k h(x) = limx->k x2 = k2

So, limx->k h(x) = h(k)

Therefore, h is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is

continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

Therefore, f(x) = (goh)(x) = cos (x2) is a continuous function.

Question 32:

Show that the function defined by f(x) = |cos x| is a continuous function.

Answer:

The given function is f(x) = |cos x|

This function f is defined for every real number and f can be written as the composition

of two functions as,

f = goh, where g(x) = |x| and h(x) = cos x

[Since (goh)(x) = g(h(x)) = g(cos x) = |cos x| = f(x)]

It has to be first proved that

g(x) = |x| and h(x) = cos x

Now, g(x) = |x| can be written as

g(x) =    -x, if x < 0

               x, if x ≥ 0

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

If c < 0, then g(c) = -c

And limx->c g(x) = limx->c (-x) = -c

So, limx->c g(x) = g(c)

Therefore, g is continuous at all points x, such that x < 0

Case II:

If c > 0, then g(c) = c

And limx->c g(x) = limx->c (x) = c

So, limx->c g(x) = g(c)

Therefore, g is continuous at all points x, such that x > 0

Case III:

If c = 0, then g(c) = g(0) = 0

limx->0- g(x) = limx->0- (-x) = 0

limx->0+ g(x) = limx->0+ (x) = 0

So, limx->0- g(x) = limx->0+ g(x) = g(0)

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h(x) = cos x

It is evident that h(x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x -> c, then h -> 0

h(c) = cos c

limx->c h(x) = limx->c cos x

                   = limh->0 cos (c + h)

                   = limh->0 [cos c * cos h - sin c * sin h]

                   = limh->0 [cos c * cos h] - limh->0 [sin c * sin h]

                   = cos c * cos 0 + sin c * sin 0

                   = cos c * 1 + sin c * 0

                   = cos c + 0

                   = cos c

So, limx->c h(x) = h(c)

Therefore, h(x) = cos x is a continuous function.

It is known that for real valued functions g and h such that (g o h) is defined at c, if g is

continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

Therefore, f(x) = (goh)(x) = g(h(x)) = g(cos x) = |cos x| is a continuous function.

Question 33:

Examine that sin |x| is a continuous function.

Answer:

Let f(x) = sin |x|

This function f is defined for every real number and f can be written as the composition of two

functions as,

f = goh, where g(x) = |x| and h(x) = sin x

Now, g(x) = |x| can be written as

g(x) =    -x, if x < 0

               x, if x ≥ 0

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

If c < 0, then g(c) = -c

And limx->c g(x) = limx->c (-x) = -c

So, limx->c g(x) = g(c)

Therefore, g is continuous at all points x, such that x < 0

Case II:

If c > 0, then g(c) = c

And limx->c g(x) = limx->c (x) = c

So, limx->c g(x) = g(c)

Therefore, g is continuous at all points x, such that x > 0

Case III:

If c = 0, then g(c) = g(0) = 0

limx->0- g(x) = limx->0- (-x) = 0

limx->0+ g(x) = limx->0+ (x) = 0

So, limx->0- g(x) = limx->0+ g(x) = g(0)

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Again, h(x) = sin x

It is evident that h(x) = sin x is defined for every real number.

Let c be a real number. Put x = c + k

If x -> c, then k -> 0

h(c) = sin c

limx->c h(x) = limx->c sin x

                   = limh->0 sin (c + h)

                   = limh->0 [sin c * cos h + cos c * sin h]

                   = limh->0 [sin c * cos h] + limh->0 [cos c * sin h]

                   = sin c * cos 0 + cos c * sin 0

                   = sin c * 1 + cos c * 0

                   = sin c + 0

                   = sin c

So, limx->c h(x) = g(c)

Therefore, h is a continuous function.

It is known that for real valued functions g and h such that (g o h) is defined at c, if g is

continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

Therefore, f(x) = sin |x| is a continuous function.

Question 34:

Find all the points of discontinuity of f defined by f(x) = |x| - |x + 1|.

Answer:

The given function is f(x) = |x| - |x + 1|

The two functions, g and h, are defined as

g(x) = |x| and h(x) = |x + 1|

Then, f = g − h

The continuity of g and h is examined first.

Now, g(x) = |x| can be written as

g(x) =    -x, if x < 0

               x, if x ≥ 0

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

If c < 0, then g(c) = -c

And limx->c g(x) = limx->c (-x) = -c

So, limx->c g(x) = g(c)

Therefore, g is continuous at all points x, such that x < 0

Case II:

If c > 0, then g(c) = c

And limx->c g(x) = limx->c (x) = c

So, limx->c g(x) = g(c)

Therefore, g is continuous at all points x, such that x > 0

Case III:

If c = 0, then g(c) = g(0) = 0

limx->0- g(x) = limx->0- (-x) = 0

limx->0+ g(x) = limx->0+ (x) = 0

So, limx->0- g(x) = limx->0+ g(x) = g(0)

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h(x) = |x + 1| can be written as

h(x) =    -(x + 1), if x < -1

               x + 1, if x ≥ -1

Clearly, h is defined for all real numbers.

Let c be a real number.

Case I:

If c < -1, then h(c) = -(c + 1)

And limx->c h(x) = limx->c [-(x + 1)] = -(c + 1)

So, limx->c h(x) = h(c)

Therefore, h is continuous at all points x, such that x < -1

Case II:

If c > -1, then h(c) = c + 1

And limx->c h(x) = limx->c (x + 1) = c + 1

So, limx->c h(x) = h(c)

Therefore, h is continuous at all points x, such that x > 0

Case III:

If c = -1, then h(c) = h(-1) = -1 + 1 = 0

limx->0- h(x) = limx->0- [-(x + 1)] = -(-1 + 1) = 0

limx->0+ h(x) = limx->0+ (x + 1) = -1 + 1 = 0

So, limx->0- h(x) = limx->0+ h(x) = h(-1)

Therefore, h is continuous at x = 0.

From the above three observations, it can be concluded that h is continuous at all points of the

real line. g and h are continuous functions. Therefore, f = g − h is also a continuous function.

Therefore, f has no point of discontinuity.

  

                                                                       Exercise 5.2

Differentiate the functions with respect to x in Exercises 1 to 8.

Question 1:

sin(x2 + 5)

Answer:

Let f(x) = sin(x2 + 5)

Differentiate w.r.t. x, we get

df(x)/dx = d{sin(x2 + 5)}/dx

                = cos(x2 + 5) * d(x2 + 5)/dx

                = cos(x2 + 5) * (2x + 0)

                = 2x cos(x2 + 5)

Question 2:

cos(sin x)

Answer:

Let f(x) = cos(sin x)

Differentiate w.r.t. x, we get

df(x)/dx = d{cos(sin x)}/dx

                = -sin(sin x) * d(sin x)/dx

                = -sin(sin x) * cos x

                = -cos x sin(sin x)

Question 3:

sin(ax + b)

Answer:

Let f(x) = sin(ax + b)

Differentiate w.r.t. x, we get

df(x)/dx = d{sin(ax + b)}/dx

                = cos(ax + b) * d(ax + b)/dx

                = cos(ax + b) * a

                = a cos(ax + b)

Question 4:

sec(tan √x)

Answer:

Let f(x) = sec(tan √x)

Differentiate w.r.t. x, we get

df(x)/dx = d{sec(tan √x)}/dx

                = sec(tan √x) * tan(tan √x) * d(tan √x)/dx

                = sec(tan √x) * tan(tan √x) * sec2 √x * d(√x)/dx

                = sec(tan √x) * tan(tan √x) * sec2 √x * (1/2√x)

                = {sec(tan √x) * tan(tan √x) * sec2 √x}/2√x

Question 5:

sin(ax + b)/cos(cx + d)

Answer:

Let f(x) = sin(ax + b)/cos(cx + d)

The given function is f(x) = sin(ax + b)/cos(cx + d) = g(x)/h(x),

where g(x) = sin(ax + b) and h(x) = cos (cx + d)

Now, f’ = (g’h – gh’)/h2

Differentiate w.r.t. x, we get

df(x)/dx = d{sin(ax + b)/cos(cx + d)}/dx

                = [a cos (ax + b).cos(cx + d) – sin(ax + b){-c * sin(cx + d)}]/{cos(cx + d)}2

                = [a cos (ax + b).cos(cx + d) + c sin(ax + b) sin(cx + d)]/{cos(cx + d)}2

                = [a cos (ax + b).cos(cx + d)]/{cos(cx + d)}2 + c sin(ax + b) sin(cx + d)]/{cos(cx + d)}2

                = a cos (ax + b).sec(cx + d) + c sin(ax + b) tan(cx + d) sec(cx + d)

Question 6:

cos x3. sin2 x5

Answer:

Let f(x) = cos x3. sin2 x5

Differentiate w.r.t. x, we get

df(x)/dx = d{cos x3. sin2 x5}/dx

               = sin2 x5 * d(cos x3)/dx + cos x3 * d(sin2 x5)/dx    

               = sin2 x5 * (-sin x3) * d(x3)/dx + cos x3 * 2sin x5 * d(sin x5)/dx

               = -sin x3 * sin2 x5 * 3x2 + 2 sin x5 * cos x3 * cos x5 * d(x5)/dx

               = -3x2 * sin x3 * sin2 x5 + 2 sin x5 * cos x3 * cos x5 * 5x4

               = 10 x4 sin x5 * cos x3 * cos x5 - 3x2 * sin x3 * sin2 x5

Question 7:

2√(cot x2)

Answer:

Let f(x) = 2√(cot x2)

Differentiate w.r.t. x, we get

df(x)/dx = d{2√(cot x2)}/dx

              = 2 * [1/2√(cot x2)] * d(cot x2)/dx

              = √[sin x2/ cos x2] * (-cosec x2) * d(x2)/dx

              = -√[sin x2/cos x2] * (1/sin x2) * (2x)

              = -(2x)/[√(sin x2)* √(cos x2) * (sin x2)]

              = -(2√2x)/[√(2 * sin x2 * cos x2) * (sin x2)]

              = -(2√2x)/[sin x2 * √(sin 2x2)]

Question 8:

 cot √x

Answer:

Let f(x) = cot √x

Differentiate w.r.t. x, we get

df(x)/dx = d(cot √x)/dx

                = -sin √x * d(√x)/dx

                = -sin √x * (1/2√x)

                = -sin √x /(2√x)        

Question 9:

Prove that the function f given by f(x) = |x - 1|, x є R is not differentiable at x = 1.

Answer:

The given function is f(x) = |x - 1|, x є R

It is known that a function f is differentiable at a point x = c in its domain if both

limh->0- {f(c + h) – f(c)}/h and limh->0+ {f(c + h) – f(c)}/h are finite and equal.

To check the differentiability of the given function at x = 1,

Consider the left hand limit of f at x = 1

   limh->0- {f(1 + h) – f(1)}/h

= limh->0- {|1 + h - 1| – |1 - 1|}/h

= limh->0- {|h| – 0}/h

= limh->0- (-h)/h                [Since h < 0 => |h| = -h]

= -1

Consider the right hand limit of f at x = 1

   limh->0+ {f(1 + h) – f(1)}/h

= limh->0+ {|1 + h - 1| – |1 - 1|}/h

= limh->0+ {|h| – 0}/h

= limh->0+ (h)/h                [Since h > 0 => |h| = h]

= 1

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1    

Question 10:

Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at     x = 1 and x = 2.

Answer:

The given function f is f(x) = [x], 0 < x < 3

It is known that a function f is differentiable at a point x = c in its domain if both

limh->0- {f(c + h) – f(c)}/h and limh->0+ {f(c + h) – f(c)}/h are finite and equal.

To check the differentiability of the given function at x = 1,

Consider the left hand limit of f at x = 1

   limh->0- {f(1 + h) – f(1)}/h

= limh->0- {[1 + h] – [1]}/h

= limh->0- (0 - 1)/h

= limh->0- (-1)/h

= ∞         

Consider the right hand limit of f at x = 1

   limh->0+ {f(1 + h) – f(1)}/h

= limh->0+ {[1 + h] – [1]}/h

= limh->0+ (1 - 1)/h

= 0         

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

To check the differentiability of the given function at x = 2,

Consider the left hand limit of f at x = 2

   limh->0- {f(2 + h) – f(2)}/h

= limh->0- {[2 + h] – [2]}/h

= limh->0- (1 - 2)/h

= limh->0- (-1)/h

= ∞  

Consider the right hand limit of f at x = 2

   limh->0+ {f(2 + h) – f(2)}/h

= limh->0+ {[2 + h] – [2]}/h

= limh->0+ (2 - 2)/h

= limh->0+ 0

= 0  

 Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2

 

                                                                       Exercise 5.3

Find dy/dx in the following:

Question 1:

2x + 3y = sin x

Answer:

Given, 2x + 3y = sin x

Differentiating w.r.t. x, we get

=> d(2x + 3y)/dx = d(sin x)/dx

=> d(2x)/dx + d(3y)/dx = d(sin x)/dx

=> 2 + 3 * dy/dx = cos x

=> 3 * dy/dx = cos x – 2

=> dy/dx = (cos x – 2)/3

Question 2:

2x + 3y = sin y

Answer:

Given, 2x + 3y = sin y

Differentiating w.r.t. x, we get

=> d(2x + 3y)/dx = d(sin y)/dx

=> d(2x)/dx + d(3y)/dx = d(sin y)/dx

=> 2 + 3 * dy/dx = cos y * dy/dx

=> 2 = cos y * dy/dx - 3 * dy/dx

=> 2 = (cos y – 3) * dy/dx

=> dy/dx = 2/(cos y – 3)

Question 3:

ax + by2 = cos y

Answer:

Given, ax + by2 = cos y

Differentiating w.r.t. x, we get

=> d(ax + by2)/dx = d(cos y)/dx

=> d(ax)/dx + d(by2)/dx = d(cos y)/dx

=> a + 2by * dy/dx = -sin y * dy/dx

=> a = -sin y * dy/dx - 2by * dy/dx

=> a = -(sin y + 2by) * dy/dx

=> dy/dx = -a/(sin y + 2by)

Question 4:

xy + y2 = tan x + y

Answer:

Given, xy + y2 = tan x + y

Differentiating w.r.t. x, we get

=> d(xy + y2)/dx = d(tan x + y)/dx

=> d(xy)/dx + d(y2)/dx = d(tan x)/dx + d(y)/dx

=> x * dy/dx  + y + 2y * dy/dx = sec2 x + dy/dx

=> x * dy/dx + 2y * dy/dx – dy/dx = sec2 x – y

=> (x + 2y - 1) * dy/dx = sec2 x – y

=> dy/dx = (sec2 x – y)/(x + 2y - 1)

 

Question 5:

x2 + xy + y2 = 100

Answer:

Given, x2 + xy + y2 = 100

Differentiating w.r.t. x, we get

=> d(x2 + xy + y2)/dx = d(100)/dx

=> d(x2)/dx + d(xy)/dx + d(y2)/dx = d(100)/dx

=> 2x + x * dy/dx + y + 2y * dy/dx = 0

=> (x + 2y) * dy/dx = -(2x + y)

=> dy/dx = -(2x + y)/(x + 2y)

Question 6:

x3 + x2y + xy2 + y3 = 81

Answer:

Given, x3 + x2y + xy2 + y3 = 81

Differentiating w.r.t. x, we get

=> d(x3 + x2y + xy2 + y3)/dx = d(81)/dx

=> d(x3)/dx + d(x2y)/dx + d(xy2)/dx + d(y3)/dx = d(81)/dx

=> 3x2 + y * d(x2)/dx + x2 * dy/dx + y2 * d(x)/dx + x * d(y2)/dx + d(y3)/dx = d(81)/dx

=> 3x2 + y * 2x + x2 * dy/dx + y2 + x * 2y * dy/dx + 3y2 * dy/dx = 0

=> 3x2 + 2xy + x2 * dy/dx + y2 + 2xy * dy/dx + 3y2 * dy/dx = 0

=> (3x2 + 2xy + y2) + (x2 + 2xy + 3y2) * dy/dx = 0

=> dy/dx = -(3x2 + 2xy + y2)/(x2 + 2xy + 3y2)

 

Question 7:

sin2 y + cos xy = π

Answer:

Given, sin2 y + cos xy = π

Differentiating w.r.t. x, we get

=> d(sin2 y + cos xy)/dx = d(π)/dx

=> d(sin2 y)/dx + d(cos xy)/dx = 0

=> 2 sin y * d(sin y)/dx – sin xy * d(xy)/dx = 0

=> 2 sin y * cos y * dy/dx – sin xy * [y * d(x)/dx + x * dy/dx] = 0

=> 2 sin y * cos y * dy/dx – sin xy * [y + x * dy/dx] = 0

=> 2 sin y * cos y * dy/dx – y * sin xy – x * sin xy * dy/dx = 0

=> 2 sin y * cos y * dy/dx – x * sin xy * dy/dx = y * sin xy

=> (2 sin y * cos y – x * sin xy) * dy/dx = y * sin xy

=> dy/dx = (y * sin xy)/(2 sin y * cos y – x * sin xy)

=> dy/dx = (y * sin xy)/(sin 2y – x * sin xy)

Question 8:

sin2 x + cos2 y = 1

Answer:

Given, sin2 x + cos2 y = 1

Differentiating w.r.t. x, we get

=> d(sin2 x + cos2 y)/dx = d(1)/dx

=> d(sin2 x)/dx + d(cos2 y)/dx = d(1)/dx

=> 2 sin x * d(sin x)/dx + 2 cos y * d(cos y)/dx = 0

=> 2 sin x * cos x - 2 cos y * sin y * dy/dx = 0

=> sin 2x - sin 2y * dy/dx = 0

=> sin 2y * dy/dx = sin 2x

=> dy/dx = sin 2x /cos 2y

Question 9:

y = sin-1{2x/(1 + x2)}

Answer:

Given, y = sin-1{2x/(1 + x2)}

=> sin y = 2x/(1 + x2)

Differentiating w.r.t. x, we get

=> d(sin y)/dx = d[2x/(1 + x2)]/dx

=> cos y * dy/dx = [(1 + x2) * d(2x)/dx – 2x * d(1 + x2)/dx]/(1 + x2)2

=> cos y * dy/dx = [2(1 + x2) – 2x * 2x]/(1 + x2)2

=> cos y * dy/dx = [2 + 2x2 – 4x2]/(1 + x2)2

=> cos y * dy/dx = [2 - 2x2]/(1 + x2)2

=> cos y * dy/dx = 2(1 - x2)/(1 + x2)2   …………1

Also, sin y = 2x/(1 + x2)

Now, cos y = √(1 – sin2 y)

=> cos y = √[1 – {2x/(1 + x2)}2]

=> cos y = √[1 – 4x2/(1 + x2)2]

=> cos y = √[{(1 + x2)2 – 4x2}/(1 + x2)2]

=> cos y = √[{(1 - x2)2/(1 + x2)2]

=> cos y = (1 - x2)/(1 + x2)

From equation 1, we get

=> (1 - x2)/(1 + x2) * dy/dx = 2(1 - x2)/(1 + x2)2   

=> dy/dx = 2/(1 + x2)   

Question 10:

y = tan-1{(3x – x3)/(1 - 3x2)}, -1/√3 < x < 1/√3

Answer:

Given, y = tan-1{(3x – x3)/(1 - 3x2)}

=> tan y = (3x – x3)/(1 - 3x2)   ……..1

It is know that tan y = (3 * tan y/3 – tan3 y/3)/(1 – 3 tan2 y/3)    …….2

Comparing equation 1 and 2, we get

x = tan y/3         ………….3

Differentiating it w.r.t. x, we get

      d(x)/dx = d(tan y/3)/dx

=> 1 = sec2 y/3 * d(y/3)/dx

=> 1 = sec2 y/3 * (1/3) * dy/dx

=> dy/dx = 3/(sec2 y/3)

=> dy/dx = 3/(1 + tan2 y/3)

=> dy/dx = 3/(1 + x2)               [From equation 3]

Question 11:

y = cos-1{(1 – x2)/(1 + x2)}, 0 < x < 1

Answer:

Given, y = cos-1{(1 – x2)/(1 + x2)}

=> cos y = (1 – x2)/(1 + x2)

=> (1 – tan2 y/2)/(1 + tan2 y/2) = (1 – x2)/(1 + x2)

On comparing LHS and RHS of the above relationship, we get

tan y/2 = x      …………..1 

Differentiating it w.r.t. x, we get

      sec2 y/2 * d(y/2)/dx = d(x)/dx

=> sec2 y/2 * (1/2) * dy/dx = 1

=> dy/dx = 2/(sec2 y/2)

=> dy/dx = 2/(1 + tan2 y/2)

=> dy/dx = 2/(1 + x2)              [From equation 1]

Question 12:

y = sin-1{(1 - x2)/(1 + x2)}

Answer:

Given, y = sin-1{(1 - x2)/(1 + x2)}

=> sin y = (1 - x2)/(1 + x2)

Differentiating w.r.t. x, we get

=> d(sin y)/dx = d[(1 - x2)/(1 + x2)]/dx  

=> cos y * dy/dx = [(1 + x2) * d(1 - x2)/dx – (1 - x2) * d(1 + x2)/dx]/(1 + x2)2

=> cos y * dy/dx = [-2x(1 + x2) – 2x(1 - x2)]/(1 + x2)2

=> cos y * dy/dx = [-2x - 2x3 – 2x + 2x3]/(1 + x2)2

=> cos y * dy/dx = -4x/(1 + x2)2    ………………..1

Also, sin y = (1 - x2)/(1 + x2)

Now, cos y = √(1 – sin2 y)

=> cos y = √[1 – {(1 - x2)/(1 + x2)}2]

=> cos y = √[1 – (1 - x2)2/(1 + x2)2]

=> cos y = √[{(1 + x2)2 – (1 - x2)2}/(1 + x2)2]

=> cos y = √[4x2/(1 + x2)2]

=> cos y = 2x/(1 + x2)

From equation 1, we get

=> 2x/(1 + x2) * dy/dx = -4x/(1 + x2)2    

=> dy/dx = -2/(1 + x2)

Question 13:

y = cos-1{2x/(1 + x2)}

Answer:

Given, y = cos-1{2x/(1 + x2)}

=> cos y = 2x/(1 + x2)

Differentiating w.r.t. x, we get

=> d(cos y)/dx = d[2x/(1 + x2)]/dx

=>-sin y * dy/dx = [(1 + x2) * d(2x)/dx – 2x * d(1 + x2)/dx]/(1 + x2)2

=> -sin y * dy/dx = [2(1 + x2) – 2x * 2x]/(1 + x2)2

=> -sin y * dy/dx = [2 + 2x2 – 4x2]/(1 + x2)2

=> -sin y * dy/dx = [2 - 2x2]/(1 + x2)2

=> -sin y * dy/dx = 2(1 - x2)/(1 + x2)2  

=> sin y * dy/dx = -2(1 - x2)/(1 + x2)2   …………1

Also, cos y = 2x/(1 + x2)

Now, sin y = √(1 – cos2 y)

=> sin y = √[1 – {2x/(1 + x2)}2]

=> sin y = √[1 – 4x2/(1 + x2)2]

=> sin y = √[{(1 + x2)2 – 4x2}/(1 + x2)2]

=> sin y = √[{(1 - x2)2/(1 + x2)2]

=> sin y = (1 - x2)/(1 + x2)

From equation 1, we get

=> (1 - x2)/(1 + x2) * dy/dx = -2(1 - x2)/(1 + x2)2   

=> dy/dx = -2/(1 + x2)

Question 14:

y = sin-1[2x√(1 – x2)], -1/√2 < x > 1/√2

Answer:

Given, y = sin-1[2x√(1 – x2)]

=> sin y = 2x√(1 – x2)

Differentiating w.r.t. x, we get

      cos y * dy/dx = 2[x * d{√(1 – x2)}/dx + √(1 – x2) * d(x)/dx]

=> cos y * dy/dx = 2[x * 1/2 * √(1 – x2) * d(1 – x2)/dx + √(1 – x2)]

=> cos y * dy/dx = 2[x * 1/2 * √(1 – x2) * (-2x) + √(1 – x2)]

=> cos y * dy/dx = 2[-x2/√(1 – x2) + √(1 – x2)]

=> cos y * dy/dx = 2[(-x2 + 1 – x2)/√(1 – x2)]

=> cos y * dy/dx = 2[(1 – 2x2)/√(1 – x2)]     ……………..1

Now, cos y = √(1 – sin2 y)

=> cos y = √[1 – {2x√(1 – x2)}2]

=> cos y = √[1 – 4x2(1 – x2)]

=> cos y = √[1 – 4x2 - 4x4]

=> cos y = √[(1 – 2x2)2]

=> cos y = (1 – 2x2)

From equation 1, we get

=> (1 – 2x2) * dy/dx = 2[(1 – 2x2)/√(1 – x2)]

=> dy/dx = 2/√(1 – x2)

Question 15:

y = sec-1{1/(2x2 – 1)}, 0 < x < 1/√2

Answer:

Given, y = sec-1{1/(2x2 – 1)}

=> sec y = 1/(2x2 – 1)

=> 1/cos y = 1/(2x2 – 1)

=> cos y = 2x2 – 1

=> 2x2 = 1 + cos y

=> 2x2 = 2 cos2 y/2

=> x2 = cos2 y/2

=> x = cos y/2

Differentiating w.r.t. x, we get

      d(x)/dx = d(cos y/2)

=> 1 = -sin y/2 * d(y/2)/dx

=> 1 = -sin y/2 * (1/2) * dy/dx

=> (1/2) * dy/dx = -1/(sin y/2)

=> dy/dx = -2/(sin y/2)

=> dy/dx = -2/√(1 – cos2 y/2)

=> dy/dx = -2/√(1 – x2)

 

                                                                       Exercise 5.4

Question 1:

Differentiate the following w.r.t. x: ex/sin x

Answer:

Let y = ex/sin x

Differentiate w.r.t. x, we get

dy/dx = d(ex/sin x)/dx

By using the quotient rule, we obtain

     dy/dx = {sin x * d(ex)/dx -  ex * d(sin x)/dx}/(sin x)2

=> dy/dx = {sin x * ex - ex * cos x}/sin2 x

=> dy/dx = ex(sin x - cos x)/sin2 x, x ≠ nπ, n є Z

Question 2:

Differentiate the following w.r.t. x: esin-1 x

Answer:

Let y = esin-1 x

Differentiate w.r.t. x, we get

dy/dx = d(esin-1 x)/dx

By using the chain rule, we get

     dy/dx = esin-1 x * d(sin-1 x)/dx

=> dy/dx = esin-1 x * 1/√(1 – x2)

=> dy/dx = esin-1 x/√(1 – x2), x є (-1, 1)

Question 3:

Differentiate the following w.r.t. x: ex3

Answer:

Let y = ex3

Differentiate w.r.t. x, we get

dy/dx = d(ex3)/dx

By using the chain rule, we obtain

     dy/dx = ex3 * d(x3)/dx

=> dy/dx = ex3 * 3x2

=> dy/dx = 3x2 ex3

Question 4:

Differentiate the following w.r.t. x: sin(tan-1 x e-x)

Answer:

Let y = sin(tan-1 e-x)

Differentiate w.r.t. x, we get

dy/dx = d{sin(tan-1 e-x)}/dx

By using the chain rule, we obtain

     dy/dx = cos(tan-1 e-x) * d(tan-1 e-x)/dx

                = cos(tan-1 e-x) * 1/{1 + (e-x)2} * d(e-x)/dx

                = cos(tan-1 e-x) * 1/{1 + (e-x)2} * (-e-x)

                = {-e-x *cos(tan-1 e-x)}/(1 + e-2x)

Question 5:

Differentiate the following w.r.t. x: log(cos ex)

Answer:

Let y = log(cos ex)

Differentiate w.r.t. x, we get

dy/dx = d{log(cos ex)}/dx

By using the chain rule, we obtain

      dy/dx = (1/cos ex) * d(cos ex)/dx

=> dy/dx = (1/cos ex) * (-sin ex) * d(ex)/dx

=> dy/dx = (1/cos ex) * (-sin ex) * ex

=> dy/dx = (-sin ex)/(cos ex) * ex

=> dy/dx = - ex * tan ex, ex ≠ (2n + 1)π/2, n є N 

Question 6:

Differentiate the following w.r.t. x: ex + ex2 + ex2 + ………+ ex5

Answer:

Let y = ex + ex2 + ex2 + ………+ ex5

Differentiate w.r.t. x, we get

      dy/dx = d(ex + ex2 + ex2 + ………+ ex5)/dx

=> dy/dx = ex + d(ex2)/dx + d(ex3)/dx + d(ex4)/dx + d(ex5)/dx

=> dy/dx = ex + ex2 * d(x2)/dx + ex3 * d(x3)/dx + ex4 * d(x4)/dx + ex5 * d(x5)/dx

=> dy/dx = ex + 2x ex2 + 3x2 ex3 + 4x3 ex4 + 5x4 ex5 

Question 7:

Differentiate the following w.r.t. x: √(e√x), x > 0

Answer:

Let y = √(e√x)

=> y2 = e√x

Differentiate w.r.t. x, we get

d(y2)/dx = d(e√x)/dx

By using the chain rule, we obtain

      2y * dy/dx = e√x * d(√x)/dx

=> 2y * dy/dx = e√x * (1/2√x)

=> dy/dx = e√x/(4y * √x)

=> dy/dx = e√x/{4 * √(e√x) * √x}

=> dy/dx = e√x/{4√(xe√x)}, x > 0

Question 8:

Differentiate the following w.r.t. x: log(log x), x > 1

Answer:

Let y = log(log x)

Differentiate w.r.t. x, we get

d(y)/dx = d{ log(log x)}/dx

By using the chain rule, we obtain

      dy/dx = (1/log x) * d(log x)/dx

=> dy/dx = (1/log x) * (1/x)

=> dy/dx = 1/(x * log x), x > 1

Question 9:

Differentiate the following w.r.t. x: cos x/log x, x > 0

Answer:

Let y = cos x/log x

Differentiate w.r.t. x, we get

dy/dx = d(cos x/log x)/dx

By using the quotient rule, we get

     dy/dx = {log x * d(cos x)/dx -  cos x * d(log x)/dx}/(log x)2

=> dy/dx = (-sin x * log x - cos x * 1/x)/(log x)2

=> dy/dx = -(x * sin x * log x + cos x)/{x * (log x)2}, x > 0

Question 10:

Differentiate the following w.r.t. x: cos(log x + ex), x > 0

Answer:

Let y = cos(log x + ex)

Differentiate w.r.t. x, we get

dy/dx = d{cos(log x + ex)}

By using the chain rule, we get

      dy/dx = -sin(log x + ex) * d[log x + ex]/dx

=> dy/dx = -sin(log x + ex) * [d(log x)/dx + d(ex)/dx]

=> dy/dx = -sin(log x + ex) * (1/x + ex)

=> dy/dx = -(1/x + ex)* sin(log x + ex), x > 0

                                                                        Exercise 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. x

Question 1:

cos x * cos 2x * cos 3x

Answer:

Let y = cos x * cos 2x * cos 3x

Taking logarithm on both the sides, we obtain

      log y = log(cos x * cos 2x * cos 3x)

=> log y = log cos x + log cos 2x + log cos 3x

Differentiating both sides with respect to x, we get

=> (1/y) * dy/dx = 1/cos x * d(cos x)/dx + 1/cos 2x * d(cos 2x)/dx + 1/cos 3x * d(cos x3)/dx

=> (1/y) * dy/dx = 1/cos x * (-sin x) + 1/cos 2x * (-sin 2x) * d(2x)/dx + 1/cos 3x * (-sin 3x) *

d(3x)/dx

=> (1/y) * dy/dx = -sin x /cos x – 2 * sin 2x /cos 2x – 3 * sin 3x/cos 3x

=> dy/dx = y[-sin x /cos x – 2 * sin 2x /cos 2x – 3 * sin 3x/cos 3x]

=> dy/dx = -cos x * cos 2x * cos 3x [tan x + 2 tan 2x + 3 tan 3x]

Question 2:

√[{(x – 1) (x – 2)}/{ (x – 3) (x – 4) (x – 5)}]

Answer:

Let y = √[{(x – 1) (x – 2)}/{ (x – 3) (x – 4) (x – 5)}]

Taking logarithm on both the sides, we obtain

      log y = log √[{(x – 1)(x – 2)}/{ (x – 3)(x – 4)(x – 5)}]

=> log y = (1/2) * log [{(x – 1)(x – 2)}/{(x – 3)(x – 4)(x – 5)}]

=> log y = (1/2) * [log{(x – 1)(x – 2)} - log{(x – 3)(x – 4)(x – 5)}]

=> log y = (1/2) * [log(x – 1) + log(x – 2) – {log(x – 3) + log(x – 4) + log(x – 5)}]

=> log y = (1/2) * [log(x – 1) + log(x – 2) – log(x – 3) - log(x – 4) - log(x – 5)]

Differentiating both sides with respect to x, we get

=> (1/y) * dy/dx = (1/2) * [1/(x – 1) * d(x - 1)/dx + 1/(x – 2) * d(x - 2)/dx – 1/(x – 3) * d(x – 3)/dx

- 1/(x – 4) * d(x - 4)/dx – 1/(x – 5) * d(x – 5)/dx]

=> (1/y) * dy/dx = (1/2) * [1/(x – 1) + 1/(x – 2) – 1/(x – 3) - 1/(x – 4) – 1/(x – 5)]

=> dy/dx = (y/2) * [1/(x – 1) + 1/(x – 2) – 1/(x – 3) - 1/(x – 4) – 1/(x – 5)]

=> dy/dx = (1/2) * √[{(x – 1) (x – 2)}/{ (x – 3) (x – 4) (x – 5)}][1/(x – 1) + 1/(x – 2) – 1/(x – 3) - 1/(x

                  – 4) – 1/(x – 5)]

Question 3:

(log x)cos x

Answer:

Let y = (log x)cos x

Taking logarithm on both the sides, we obtain

      log y = log[(log x)cos x]

=> log y = cos x * log(log x)

Differentiating both sides with respect to x, we get

=> (1/y) * dy/dx = d[cos x * log(log x)]/dx

=> (1/y) * dy/dx = cos x * d[log(log x)]/dx + log(log x) * d(cos x)/dx

=> (1/y) * dy/dx = cos x * (1/log x) * d(log x)/dx + log(log x) * (-sin x)

=> (1/y) * dy/dx = cos x * (1/log x) * (1/x) – sin x * log(log x)

=> dy/dx = y[cos x /(x * log x) – sin x * log(log x)]

=> dy/dx = (log x)cos x[cos x /(x * log x) – sin x * log(log x)]

Question 4:

xx – 2sin x

Answer:

Let y = xx – 2sin x

Taking logarithm on both the sides, we obtain

      log y = log(xx – 2sin x)

=> log y = log(xx) – log(2sin x)

=> log y = x * log x – sin x * log 2

Differentiating both sides with respect to x, we

=> (1/y) * dy/dx = d[x * log x – sin x * log 2]/dx

=> (1/y) * dy/dx = d[x * log x]/dx – d[sin x * log 2]/dx

=> (1/y) * dy/dx = [x * d(log x)/dx + log x * d(x)/dx] – log 2 * d(sin x)/dx

=> (1/y) * dy/dx = [x * (1/x) + log x] – log 2 * cos x

=> (1/y) * dy/dx = 1 + log x – log 2 * cos x

=> dy/dx = y[1 + log x – log 2 * cos x]

=> dy/dx = (xx – 2sin x)[1 + log x – log 2 * cos x]

Question 5:

(x + 3)2(x + 4)3(x + 5)4

Answer:

Let y = (x + 3)2(x + 4)3(x + 5)4

Taking logarithm on both the sides, we obtain

      log y = log[(x + 3)2(x + 4)3(x + 5)4]

=> log y = log(x + 3)2 + log(x + 4)3 + log(x + 5)4

=> log y = 2 * log(x + 3) + 3 * log(x + 4) + 4 * log(x + 5)

Differentiating both sides with respect to x, we

=> (1/y) * dy/dx = 2/(x + 3) * d(x + 3)/dx + 3/(x + 4) * d(x + 4)/dx + 4/(x + 5) * d(x + 5)/dx

=> (1/y) * dy/dx = 2/(x + 3) + 3/(x + 4) + 4/(x + 5)

=> dy/dx = y[2/(x + 3) + 3/(x + 4) + 4/(x + 5)]

=> dy/dx = y[{2(x + 4)(x + 5) + 3(x + 3)(x + 5) + 4(x + 3)(x + 4)}/{(x + 3)(x + 4)(x + 5)}]

=> dy/dx = y[{2(x2 + 9x + 20) + 3(x2 + 8x + 15) + 4(x2 + 7x + 12)}/{(x + 3)(x + 4)(x + 5)}]

=> dy/dx = y[{2x2 + 18x + 40 + 3x2 + 24x + 45 + 4x2 + 28x + 48}/{(x + 3)(x + 4)(x + 5)}]

=> dy/dx = y[{9x2 + 70x + 133}/{(x + 3)(x + 4)(x + 5)}]

=> dy/dx = (x + 3)2(x + 4)3(x + 5)4[{9x2 + 70x + 133}/{(x + 3)(x + 4)(x + 5)}]

=> dy/dx = (x + 3)(x + 4)2(x + 5)3(9x2 + 70x + 133)

Question 6:

(x + 1/x)x + x(1 + 1/x)

Answer:

Let y = (x + 1/x)x + x(1 + 1/x)

Taking logarithm on both the sides, we obtain

      log y = log[(x + 1/x)x + x(1 + 1/x)]

=> log y = log[(x + 1/x)x] + log[x(1 + 1/x)]

=> log y = x * log (x + 1/x) + (1 + 1/x) * log x

Differentiating both sides with respect to x, we

=> (1/y) * dy/dx = d[x * log (x + 1/x) + (1 + 1/x) * log x]/dx

=> (1/y) * dy/dx = d[x * log (x + 1/x)]/dx + d[(1 + 1/x) * log x]/dx

=> (1/y) * dy/dx = x * d[log (x + 1/x)]/dx + log (x + 1/x) * d(x)/dx + (1 + 1/x) * d[log x]/dx + log x

                                 * d[(1 + 1/x)]/dx

=> (1/y) * dy/dx = x * 1/(x + 1/x) * d(x + 1/x)/dx + log (x + 1/x) + (1 + 1/x) * (1/x) + log x * (1/x2)                              

=> (1/y) * dy/dx = x/(x + 1/x) * (1 - 1/x2) + log (x + 1/x) + (1 + 1/x) * (1/x) + log x /x2

=> (1/y) * dy/dx = x/{{x2 + 1)/x} * {(x2 – 1)/x2} + log (x + 1/x) + (1 + 1/x) * (1/x) + log x /x2

=> (1/y) * dy/dx = {x2/(x2 + 1)} * {(x2 – 1)/x2} + log (x + 1/x) + (1 + 1/x) * (1/x) + log x /x2

=> (1/y) * dy/dx = {(x2 – 1)/ (x2 + 1)} + log (x + 1/x) + (1 + 1/x) * (1/x) + log x /x2

=> dy/dx = y[{(x2 – 1)/ (x2 + 1)} + log (x + 1/x) + (1 + 1/x) * (1/x) + log x /x2]

=> dy/dx = [(x + 1/x)x + x(1 + 1/x)][{(x2 – 1)/ (x2 + 1)} + log (x + 1/x) + (1 + 1/x) * (1/x) + log x /x2]

Question 7:

(log x)x + xlog x

Answer:

Let y = (log x)x + xlog x

Again, let u = (log x)x and v = xlog x

So, y = u + v

=> dy/dx = du/dx + dv/dx   ……….1

Now, u = (log x)x

Taking log on both sides, we get

     log u = log[(log x)x]

=> log u = x * log(log x)

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = d[x * log(log x)]/dx

=> (1/u) * du/dx = d(x)/dx * log(log x) + x * d[log(log x)]/dx

=> (1/u) * du/dx = log(log x) + x * (1/log x) * d(log x)/dx

=> (1/u) * du/dx = log(log x) + x * (1/log x) * (1/x)

=> (1/u) * du/dx = log(log x) + 1/log x

=> (1/u) * du/dx = [log(log x) * log x + 1]/log x

=> du/dx = u[log(log x) * log x + 1]/log x

=> du/dx = (log x)x[log(log x) * log x + 1]/log x

=> du/dx = (log x)x-1[log(log x) * log x + 1]   ……….2

Again, v = xlog x

Taking log on both sides, we get

      log v = log[xlog x]

=> log v = log x * log x

=> log v = (log x)2

Differentiate w.r.t. x, we get

=> (1/v) * dv/dx = d[(log x)2]/dx

=> (1/v) * dv/dx = 2 * log x * d(log x)/dx

=> (1/v) * dv/dx = 2 * log x * 1/x

=> (1/v) * dv/dx = (2 * log x)/x

=> dv/dx = (2v * log x)/x

=> dv/dx = (2xlog x * log x)/x

=> dv/dx = 2xlog x - 1 * log x    ………3

From equation 1, 2 and 3, we get

=> dy/dx = (log x)x-1[log(log x) * log x + 1] + 2xlog x - 1 * log x

 

Question 8:

(sin x)x + sin-1(√x)

Answer:

Let y = (sin x)x + sin-1(√x)

Again, let u = (sin x)x and v = sin-1(√x)

So, y = u + v

=> dy/dx = du/dx + dv/dx   ……….1

Now, u = (sin x)x

Taking log on both sides, we get

     log u = log[(sin x)x]

=> log u = x * log(sin x)

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = d[x * log(sin x)]/dx

=> (1/u) * du/dx = d(x)/dx * log(sin x) + x * d[log(sin x)]/dx

=> (1/u) * du/dx = log(sin x) + x * (1/sin x) * d(sin x)/dx

=> (1/u) * du/dx = log(sin x) + x * (cos x /sin x)

=> (1/u) * du/dx = log(sin x) + x * cot x

=> du/dx = u[log(sin x) + x * cot x]

=> du/dx = (sin x)x[log(sin x) + x * cot x]    ………………2

Again, v = sin-1 (√x)

Taking log on both sides, we get

Differentiate w.r.t. x, we get

=> (1/v) * dv/dx = d[sin-1 (√x)]/dx

=> (1/v) * dv/dx = [1/√{1 – (√x)2}] * d(√x)/dx

=> (1/v) * dv/dx = [1/√(1 – x)] * (1/2√x)

=> (1/v) * dv/dx = 1/2√(x – x2)    ……………3

From equation 1, 2 and 3, we get

=> dy/dx = (sin x)x[log(sin x) + x * cot x] + 1/2√(x – x2)

Question 9:

xsin x + (sin x)cos x

Answer:

Let y = xsin x + (sin x)cos x

Again, let u = xsin x and v = (sin x)cos x

So, y = u + v

=> dy/dx = du/dx + dv/dx   ……….1

Now, u = xsin x

Taking log on both sides, we get

     log u = log[xsin x]

=> log u = sin x * log x

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = d[sin x * log x]/dx

=> (1/u) * du/dx = d(sin x)/dx * log x + sin x * d(log x)/dx

=> (1/u) * du/dx = cos x * log x + sin x /x

=> du/dx = u[cos x * log x + sin x /x]

=> du/dx = xsin x[cos x * log x + sin x /x]    ………2

Again, v = (sin x)cos x

Taking log on both sides, we get

      log v = log[(sin x)cos x]

=> log v = cos x * log(sin x)

Differentiate w.r.t. x, we get

=> (1/v) * dv/dx = d[cos x * log(sin x)]/dx

=> (1/v) * dv/dx = d(cos x)/dx * log(sin x) + cos x * d[log(sin x)]/dx

=> (1/v) * dv/dx = (-sin x) * log(sin x) + cos x * (1/sin x) * d(sin x)/dx

=> (1/v) * dv/dx = (-sin x) * log(sin x) + cos x * (1/sin x) * cos x

=> (1/v) * dv/dx = (-sin x) * log(sin x) + (cos x /sin x) * cos x

=> (1/v) * dv/dx = -sin x * log(sin x) + cot x * cos x

=> dv/dx = v[-sin x * log(sin x) + cot x * cos x]

=> dv/dx = (sin x)cos x[-sin x * log(sin x) + cot x * cos x]    …………3

From equation 1, 2 and 3, we get

=> dy/dx = xsin x[cos x * log x + sin x /x] + (sin x)cos x[-sin x * log(sin x) + cot x * cos x]

Question 10:

xx * cos x + (x2 + 1)/(x2 - 1)

Answer:

Let y = xx * cos x + (x2 + 1)/(x2 - 1)

Again, let u = xx * cos x and v = (x2 + 1)/(x2 - 1)

So, y = u + v

=> dy/dx = du/dx + dv/dx   ……….1

Now, u = xx * cos x

Taking log on both sides, we get

     log u = log[xx * cos x]

=> log u = x * cos x * log x

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = d[x * cos x * log x]/dx

=> (1/u) * du/dx = d(x)/dx * cos x * log x + x * d(cos x * log x)/dx

=> (1/u) * du/dx = d(x)/dx * cos x * log x + x * [d(cos x)/dx * log x + cos x * d(log x)/dx]

=> (1/u) * du/dx = cos x * log x + x * [(-sin x) * log x + cos x /x]

=> (1/u) * du/dx = cos x * log x - x * sin x * log x + cos x

=> du/dx = u[cos x * log x - x * sin x * log x + cos x]

=> du/dx = xx * cos x[cos x * log x - x * sin x * log x + cos x]   ……..2

Again, v = (x2 + 1)/(x2 - 1)

Taking log on both sides, we get

      log v = log[(x2 + 1)/(x2 - 1)]

=> log v = log(x2 + 1) - log(x2 - 1)

Differentiate w.r.t. x, we get

=> (1/v) * dv/dx = d[log(x2 + 1) - log(x2 - 1)]/dx

=> (1/v) * dv/dx = 1/(x2 + 1) * d(x2 + 1)/dx - 1/(x2 - 1) * d(x2 - 1)/dx

=> (1/v) * dv/dx = 2x/(x2 + 1) - 2x/(x2 - 1)

=> (1/v) * dv/dx = [2x(x2 - 1) - 2x(x2 + 1)]/[(x2 - 1)(x2 + 1)]

=> (1/v) * dv/dx = (2x3 - 2x - 2x3 – 2x)/[(x2 - 1)(x2 + 1)]

=> (1/v) * dv/dx = (-4x)/[(x2 - 1)(x2 + 1)]

=> dv/dx = v * (-4x)/[(x2 - 1)(x2 + 1)]

=> dv/dx = [(x2 + 1)/(x2 - 1)] * (-4x)/[(x2 - 1)(x2 + 1)]

=> dv/dx = -4x/(x2 - 1)2   …………..3

From equation 1, 2 and 3, we get

=> dy/dx = xx * cos x[cos x * log x - x * sin x * log x + cos x] - 4x/(x2 - 1)2

Question 11:

(x * cos x)x + (x * sin x)1/x

Answer:

Let y = (x * cos x)x + (x * sin x)1/x

Again, let u = (x * cos x)x and v = (x * sin x)1/x

So, y = u + v

=> dy/dx = du/dx + dv/dx   ……….1

Now, u = (x * cos x)x

Taking log on both sides, we get

     log u = log[(x * cos x)x]

=> log u = x * log(x * cos x)

=> log u = x * [log x + log(cos x)]

=> log u = x * log x + x * log(cos x)

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = d[x * log x + x * log(cos x)]/dx

=> (1/u) * du/dx = d[x * log x]/dx + d[x * log(cos x)]/dx

=> (1/u) * du/dx = d(x)/dx * log x + x * d(log x)/dx + d(x)/dx * log(cos x) + x * d[log(cos x)]/dx

=> (1/u) * du/dx = log x + x * (1/x) + log(cos x) + x * (1/cos x) * d(cos x)/dx

=> (1/u) * du/dx = log x + 1 + log(cos x) + x * (1/cos x) * (-sin x)

=> (1/u) * du/dx = log x + 1 + log(cos x) - x * tan x

=> (1/u) * du/dx = 1 - x * tan x + log x + log(cos x)

=> (1/u) * du/dx = 1 - x * tan x + log(x * cos x)

=> du/dx = u[1 - x * tan x + log(x * cos x)]

=> du/dx = (x * cos x)x[1 - x * tan x + log(x * cos x)]    ……………..2

Again, v = (x * sin x)1/x

Taking log on both sides, we get

      log v = log[(x * sin x)1/x]

=> log v = (1/x) * log[(x * sin x)]

=> log v = (1/x) * [log x + log(sin x)]

=> log v = log x /x + log(sin x) /x

Differentiate w.r.t. x, we get

=> (1/v) * dv/dx = d[log x /x + log(sin x) /x]/dx

=> (1/v) * dv/dx = d[log x /x]/dx + d[log(sin x) /x]/dx

=> (1/v) * dv/dx = [(1/x) * d(log x) + log x * d(1/x)/dx]+ [log(sin x) * d(1/x)/dx + (1/x) * d[log(sin

                                 x)]/dx]

=> (1/v) * dv/dx = [(1/x) * (1/x) + log x * (-1/x2) + [log(sin x) * (-1/x2) + (1/x) * (1/sin x) * d(sin

                                  x)/dx                            

=> (1/v) * dv/dx = [1/x2 - log x /x2] + [-log(sin x) /x2 + (1/x) * (1/sin x) * cos x]

=> (1/v) * dv/dx = [1/x2 - log x /x2] + [-log(sin x) /x2 + cos x /x]

=> (1/v) * dv/dx = (1 - log x)/x2 + [-log(sin x) /x2 + cos x /x]

=> (1/v) * dv/dx = (1 - log x)/x2 + [{-log(sin x) + x * cot x}/x2]

=> (1/v) * dv/dx = [1 - log x - log(sin x) + x * cot x]/x2

=> (1/v) * dv/dx = [1 – {log x + log(sin x)} + x * cot x]/x2

=> (1/v) * dv/dx = [1 –  log(x * sin x) + x * cot x]/x2

=> dv/dx = v[1 –  log(x * sin x) + x * cot x]/x2

=> dv/dx = (x * sin x)1/x[1 –  log(x * sin x) + x * cot x]/x2    ……….3

From equation 1, 2 and 3, we get

=> dy/dx = (x * cos x)x[1 - x * tan x + log(x * cos x)] + (x * sin x)1/x[1 –  log(x * sin x) + x * cot x]/x2

Find dy/dx of the functions given in Exercises 12 to 15.

Question 12:

xy + yx = 1

Answer:

Given, xy + yx = 1

Again, let u = xy and v = yx

So, u + v = 1

=> du/dx + dv/dx = 0   ……….1

Now, u = xy

Taking log on both sides, we get

     log u = log[xy]

=> log u = y * log x

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = d[y * log x]/dx

=> (1/u) * du/dx = d(y)/dx * log x + y * d(log x)/dx

=> (1/u) * du/dx = log x * dy/dx + y/x

=> du/dx = u[log x * dy/dx + y/x]

=> du/dx = xy[log x * dy/dx + y/x]   ……………..2

Again, v = yx

Taking log on both sides, we get

     log v = log[yx]

=> log v = x * log y

Differentiate w.r.t. x, we get

=> (1/v) * dv/dx = d[x * log y]/dx

=> (1/v) * dv/dx = d(x)/dx * log y + x * d(log y)/dx

=> (1/yx) * dv/dx = log y + (x/y) * dy/dx

=> dv/dx = yx[log y + (x/y) * dy/dx]   ……………..3

From equation 1, 2 and 3, we get

=> xy[log x * dy/dx + y/x] + yx[log y + (x/y) * dy/dx] = 0

=> xy * log x * dy/dx + yxy-1 + yx * log y + xyx-1 * dy/dx = 0

=> [xy * log x + xyx-1]* (dy/dx) = -[ yxy-1 + yx * log y]

=> dy/dx = -[yxy-1 + yx * log y]/[xy * log x + xyx-1]

Question 13:

yx = xy

Answer:

Given, yx = xy

Taking log on both sides, we get

      log(yx)= log(xy)

=> x * log y = y * log x

Differentiate w.r.t. x, we get

=> d[x * log y]/dx = d[y * log x]/dx

=> d(x)/dx * log y + x * d(log y)/dx = d(y)/dx * log x + y * d(log x)/dx

=> log y + (x/y) * dy/dx = dy/dx * log x + y/x

=> (x/y) * dy/dx - dy/dx * log x = y/x – log y

=> (x/y – log x) * dy/dx = y/x – log y

=> [(x – y * log x)/y] * dy/dx = [(y – x * log y)]/x

=> x(x – y * log x) * dy/dx = y(y – x * log y)

=> dy/dx = [y(y – x * log y)]/[x(x – y * log x)]

Question 14:

(cos x)y = (cos y)x

Answer:

Given, (cos x)y = (cos y)x

Taking log on both sides, we get

      log (cos x)y = log (cos y)x

=> y * log (cos x) = x * log (cos y)

Differentiate w.r.t. x, we get

=> d[y * log (cos x)]/dx = d[x * log (cos y)]/dx

=> dy/dx * log (cos x) + y * d[log (cos x)]/dx = d(x)/dx * log (cos y) + x * d[log (cos y)]/dx

=> dy/dx * log (cos x) + (y/cos x) * d(cos x)/dx = log (cos y) + (x/cos y) * d(cos y)/dx

=> dy/dx * log (cos x) + (y/cos x) * (-sin x) = log (cos y) + (x/cos y) * (-sin y) * dy/dx

=> dy/dx * log (cos x) - (y * sin x)/cos x = log (cos y) - (x * sin y)/cos y * dy/dx

=> dy/dx * log (cos x) + (x * sin y)/cos y * dy/dx = log (cos y) + (y * sin x)/cos x 

=> dy/dx * log (cos x) + x * tan y * dy/dx = log (cos y) + y * tan x 

=> [log (cos x) + x * tan y] * dy/dx = log (cos y) + y * tan x

=> dy/dx = [log (cos y) + y * tan x]/[log (cos x) + x * tan y]

Question 15:

xy = e(x - y)

Answer:

Given, xy = e(x - y)

Taking log on both sides, we get

      log (xy) = log [e(x - y)]

=> log x + log y = x - y

Differentiate w.r.t. x, we get

=> d[log x + log y]/dx = d[x - y]/dx

=> d(log x)/dx + d(log y)/dx = d(x)/dx – d(y)/dx

=> 1/x + 1/y * dy/dx = 1 – dy/dx

=> 1/y * dy/dx + dy/dx = 1 – 1/x

=> (1/y + 1) * dy/dx = (x – 1)/x

=> {(1 + y)/y} * dy/dx = (x – 1)/x

=> x(1 + y) * dy/dx = y(x – 1)

=> dy/dx = [y(x – 1)]/[x(y + 1)]

Question 16:

Find the derivative of the function given by f(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8) and hence find f’(1).

Answer:

The given relationship is f(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8)

Taking logarithm on both the sides, we obtain

      log[f(x)] = log[(1 + x)(1 + x2)(1 + x4)(1 + x8)]

=> log[f(x)] = log(1 + x) + log(1 + x2) + log(1 + x4) + log(1 + x8)

Differentiating both sides with respect to x, we obtain

=> d[log{f(x)}]/dx = d[log(1 + x)]/dx + d[log(1 + x2)]/dx + d[log(1 + x4)]/dx + d[log(1 + x8)]/dx

=> {1/f(x)} * d[f(x)]/dx = [1/(1 + x)] * d(1 + x)/dx + [1/(1 + x2)] * d(1 + x2)/dx + [1/(1 + x4)] * d(1 +

                                             x4)/dx + [1/(1 + x8)] * d(1 + x8)/dx

=> {1/f(x)} * f’(x) = [1/(1 + x)] + [1/(1 + x2)] * 2x + [1/(1 + x4)] * 4x3 + [1/(1 + x8)] * 8x7

=> f’(x) = f(x)[1/(1 + x) + {1/(1 + x2)} * 2x + {1/(1 + x4)} * 4x3 + {1/(1 + x8)} * 8x7]

=> f’(x) = f(x)[1/(1 + x) + 2x/(1 + x2) + 4x3/(1 + x4) + 8x7/(1 + x8)]

=> f’(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8)[1/(1 + x) + 2x/(1 + x2) + 4x3/(1 + x4) + 8x7/(1 + x8)]

Now, f’(x) = (1 + 1)(1 + 12)(1 + 14)(1 + 18)[1/(1 + 1) + 2/(1 + 12) + 4/(1 + 14) + 8/(1 + 18)]

=> f’(x) = 2 * 2 * 2 * 2 * [1/2 + 2/2 + 4/2 + 8/2]

=> f’(x) = 16 * (1 + 2 + 4 + 8)/2

=> f’(x) = 8 * 15

=> f’(x) = 120

Question 17:

Differentiate (x2 – 5x + 8)(x3 + 7x + 9) in three ways mentioned below

(i) By using product rule.               

(ii) By expanding the product to obtain a single polynomial.

(iii By logarithmic differentiation.

Do they all give the same answer?

Answer:

Let y = (x2 – 5x + 8)(x3 + 7x + 9)

(i) Let u = x2 – 5x + 8 and v = x3 + 7x + 9

So, y = uv

Now, dy/dx = (du/dx) * v + u * (dv/dx)

=> dy/dx = d(x2 – 5x + 8)/dx * (x3 + 7x + 9) + (x2 – 5x + 8) * d(x3 + 7x + 9)/dx

=> dy/dx = (2x - 5) * (x3 + 7x + 9) + (x2 – 5x + 8) * (3x2 + 7)

=> dy/dx = 2x(x3 + 7x + 9) - 5(x3 + 7x + 9) + 3x2(x2 – 5x + 8) + 7(x2 – 5x + 8)

=> dy/dx = 2x4 + 14x2 + 18x - 5x3 - 35x - 45 + 3x4 – 15x3 + 24x2 + 7x2 – 35x + 56

=> dy/dx = 5x4 - 20x3 + 45x2 - 52x + 11

(ii) y = (x2 – 5x + 8)(x3 + 7x + 9)

=> y = x2(x3 + 7x + 9) – 5x(x3 + 7x + 9) + 8(x3 + 7x + 9)

=> y = x5 + 7x3 + 9x2 – 5x4 - 35x2 – 45x + 8x3 + 56x + 72

=> y = x5 - 5x4 + 15x3 - 26x2 + 11x + 72

Differentiating both sides with respect to x, we get

=> dy/dx = d[x5 - 5x4 + 15x3 - 26x2 + 11x + 72]/dx

=> dy/dx = d(x5)/dx – d(5x4)/dx + d(15x3)/dx – d(26x2)/dx + d(11x)/dx + d(72)/dx

=> dy/dx = 5x4 – 5 * 4x3 + 15 * 3x2 – 26 * 2x + 11

=> dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11

(iii) y = (x2 – 5x + 8)(x3 + 7x + 9)

Taking log on both sides, we get

      log y = log[(x2 – 5x + 8)(x3 + 7x + 9)]

=> log y = log(x2 – 5x + 8) + log(x3 + 7x + 9)

Differentiating both sides with respect to x, we get

=> 1/y * dy/dx = 1/(x2 – 5x + 8) * d(x2 – 5x + 8)/dx + 1/(x3 + 7x + 9) * d(x3 + 7x + 9)/dx

=> 1/y * dy/dx = 1/(x2 – 5x + 8) * (2x - 5) + 1/(x3 + 7x + 9) * (3x2 + 7)

=> 1/y * dy/dx = (2x - 5)/(x2 – 5x + 8) + (3x2 + 7)/(x3 + 7x + 9)

=> 1/y * dy/dx = [(2x - 5)(x3 + 7x + 9) + (3x2 + 7)(x2 – 5x + 8)]/[(x2 – 5x + 8)(x3 + 7x + 9)]

=> dy/dx = y[(2x - 5)(x3 + 7x + 9) + (3x2 + 7)(x2 – 5x + 8)]/[(x2 – 5x + 8)(x3 + 7x + 9)]

=> dy/dx = (x2 – 5x + 8)(x3 + 7x + 9)[(2x - 5)(x3 + 7x + 9) + (3x2 + 7)(x2 – 5x + 8)]/[(x2 – 5x + 8)(x3 +

                    7x + 9)]

=> dy/dx = (2x - 5)(x3 + 7x + 9) + (3x2 + 7)(x2 – 5x + 8)

=> dy/dx = 2x(x3 + 7x + 9) - 5(x3 + 7x + 9) + 3x2(x2 – 5x + 8) + 7(x2 – 5x + 8)

=> dy/dx = 2x4 + 14x2 + 18x - 5x3 - 35x - 45 + 3x4 – 15x3 + 24x2 + 7x2 – 35x + 56

=> dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11

From the above three observations, it can be concluded that all the results of dy/dx are same.

Question 18:

If u, v and w are functions of x, then show that

d(u.v.w)/dx = (du/dx).v.w + u.(du/dx).w + u.v.(dw/dx)

in two ways-first by repeated application of product rule, second by logarithmic differentiation.

Answer:

Let y = u.v.x = u.(v.w)

By applying product rule, we get

     dy/dx = du/dx . (v.w) + u . d(v.w)/dx

=> dy/dx = du/dx . (v.w) + u . [(dv/dx).w + v. (dw/dx)]

=> dy/dx = du/dx . (v.w) + u . (dv/dx).w + u.v. (dw/dx)

By taking logarithm on both sides of the equation y = u.v.w, we get

      log y = log(u.v.w)

=> log y = log u + log v + log w

Differentiating both sides with respect to x, we obtain

=> 1/y * dy/dx = d(log u)/dx + d(log v)/dx + d(log w)/dx

=> 1/y * dy/dx = 1/u * du/dx + 1/v * dv/dx + 1/w * dw/dx

=> dy/dx = y[1/u * du/dx + 1/v * dv/dx + 1/w * dw/dx]

=> dy/dx = u.v.w.[1/u * du/dx + 1/v * dv/dx + 1/w * dw/dx]

=> dy/dx = du/dx . (v.w) + u . (dv/dx).w + u.v. (dw/dx)

                                                                        Exercise 5.6

If x and y are connected parametrically by the equations given in Exercises 1 to 10,

without eliminating the parameter, Find dy/dx.

Question 1:

x = 2at2, y = at4

Answer:

The given equations are: x = 2at2, y = at4

Now, dx/dt = d(2at2)/dt

=> dx/dt = 2a * d(t2)/dt

=> dx/dt = 2a * 2t

=> dx/dt = 4at

and dy/dt = d(at4)/dt

=> dx/dt = a * d(t4)/dt

=> dx/dt = a * 4t3

=> dx/dt = 4at3

So, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = 4at3/4at

=> dy/dx = t2 

Question 2:

x = a cos θ, y = b cos θ

Answer:

The given equations are: x = a cos θ, y = b cos θ

Now, dx/dθ = d(a cos θ)/dθ

=> dx/dθ = a * d(cos θ)/dθ

=> dx/dθ = -a sin θ

and dy/dθ = d(b cos θ)/dθ

=> dy/dθ = b * d(cos θ)/dθ

=> dy/dθ = -b sin θ

Now, dy/dx = (dy/dθ)/(dx/dθ)

=> dy/dx = (-b sin θ)/(-a sin θ)

=> dy/dx = b/a

Question 3:

x = sin t, y = cos 2t

Answer:

The given equations are: x = sin t, y = b cos 2t

Now, dx/dt = d(sin t)/dt

=> dx/dt = cos t

and dy/dt = d(cos 2t)/dt

=> dy/dt = -sin 2t * d(2t)/dt

=> dy/dt = -2 sin 2t

Now, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = (-2 sin 2t)/cos t

=> dy/dx = (-2 * 2 * sin t * cos t)/cos t

=> dy/dx = -4 sin t

Question 4:

x = 4t, y = 4/t

Answer:

The given equations are: x = 4t, y = 4/t

Now, dx/dt = d(4t)/dt

=> dx/dt = 4

and dy/dt = d(4/t)/dt

=> dy/dt = -4/t2

Now, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = (-4/t2)/4

=> dy/dx = - 1/t2

Question 5:

x = cos θ - cos 2θ, y = sin θ - sin 2θ

Answer:

The given equations are: x = cos θ - cos 2θ, y = sin θ - sin 2θ

Now, dx/dθ = d(cos θ - cos 2θ)/dθ

=> dx/dθ = d(cos θ)/dθ - d(cos 2θ)/dθ

=> dx/dθ = -sin θ – (-2sin 2θ)

=> dx/dθ = 2sin 2θ - sin θ

and, dy/dθ = d(sin θ - sin 2θ)/dθ

=> dy/dθ = d(sin θ)/dθ - d(sin 2θ)/dθ

=> dy/dθ = cos θ – 2cos 2θ

Now, dy/dx = (dy/dθ)/(dx/dθ)

=> dy/dx = (cos θ – 2cos 2θ)/(2sin 2θ - sin θ)

Question 6:

x = a(θ - sin θ), y = a(1 + cos θ)

Answer:

The given equations are: x = a(θ - sin θ), y = a(1 + cos θ)

Now, dx/dθ = d{a(θ - sin θ)}/dθ

=> dx/dθ = a[d(θ)/dθ - d(sin θ)/dθ]

=> dx/dθ = a(1 – cos θ)

and, dy/dθ = d{a(1 + cos θ)}/dθ

=> dy/dθ = a[d(1)/dθ + d(cos θ)/dθ]

=> dy/dθ = a(0 - sin θ)

=> dy/dθ = -a sin θ

Now, dy/dx = (dy/dθ)/(dx/dθ)

=> dy/dx = (-a sin θ)/{ a(1 – cos θ)}

=> dy/dx = -sin θ/(1 – cos θ)

=> dy/dx = (-2 * sin θ/2 * cos θ/2)/(2 sin2 θ/2)

=> dy/dx = (-cos θ/2)/(sin θ/2)

=> dy/dx = -cot θ/2

Question 7:

x = sin3 t/√(cos 2t), y = cos3 t/√(cos 2t)

Answer:

The given equations are: x = sin3 t/√(cos 2t), y = cos3 t/√(cos 2t)

Now, dx/dt = d{sin3 t/√(cos 2t)}/dt

=> dx/dt = {√(cos 2t) * d(sin3 t)/dt - sin3 t *  d(√(cos 2t)/dt}/{√(cos 2t)}2

=> dx/dt = {√(cos 2t) * 3 sin2 t * d(sin t)/dt - sin3 t * (1/2√(cos 2t) * d(cos 2t)/dt}/cos 2t

=> dx/dt = {3√(cos 2t) * sin2 t * cos t - sin3 t * (1/2√(cos 2t) * (-2 sin 2t)}/cos 2t

=> dx/dt = {3√(cos 2t) * sin2 t * cos t - sin3 t * (1/2√(cos 2t) * (-2 sin 2t)}/cos 2t

=> dx/dt = (3 cos 2t * sin2 t * cos t + sin3 t * sin 2t)/{cos 2t * √(cos 2t)}

and dy/dt = d{cos3 t/√(cos 2t)}/dt

=> dy/dt = {√(cos 2t) * d(cos3 t)/dt - cos3 t *  d(√(cos 2t)/dt}/{√(cos 2t)}2

=> dy/dt = {√(cos 2t) * 3 cos2 t * d(cos t)/dt - cos3 t * (1/2√(cos 2t) * d(cos 2t)/dt}/cos 2t

=> dy/dt = {3√(cos 2t) * cos2 t * (-sin t) - cos3 t * (1/2√(cos 2t) * (-2 sin 2t)}/cos 2t

=> dy/dt = (-3 cos 2t * cos2 t * sin t + cos3 t * sin 2t)/{cos 2t * √(cos 2t)}

Now, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = (-a sin θ)/{ a(1 – cos θ)}

=> dy/dx = [(-3 cos 2t * cos2 t * sin t + cos3 t * sin 2t)/{cos 2t * √(cos 2t)}]

                    [(3 cos 2t * sin2 t * cos t + sin3 t * sin 2t)/{cos 2t * √(cos 2t)}]

=> dy/dx = [(-3 cos 2t * cos2 t * sin t + cos3 t * sin 2t

                    [(3 cos 2t * sin2 t * cos t + sin3 t * sin 2t)

=> dy/dx = [(-3 cos 2t * cos2 t * sin t + cos3 t * 2 * sin t * cos t]

                    [(3 cos 2t * sin2 t * cos t + sin3 t * 2 * sin t * cos t]

=> dy/dx = [sin t * cos t (-3 cos 2t * cos t + 2 * cos3 t)]

                    [sin t * cos t (3 cos 2t * sin t + 2 * sin3 t)]

=> dy/dx = [-3 cos 2t * cos t + 2 * cos3 t]/[3 cos 2t * sin t + 2 * sin3 t]

=> dy/dx = [-3(2 cos2 t - 1) * cos t + 2 * cos3 t]/[3(2 cos2 t - 1) * sin t + 2 * sin3 t]

=> dy/dx = [-4 cos3 t + 3 cos t]/[3 sin t - 4 sin3 t] 

=> dy/dx = [-cos 3t]/[sin 3t]                      [cos 3t = 4 cos3 t - 3 cos t and sin 3t = 3 sin t - 4 sin3 t]

=> dy/dx = -cot 3t

Question 8:

x = a(cos t + log tan t/2), y = a sin t

Answer:

The given equations are: x = a(cos t + log tan t/2), y = a sin t

Now, dx/dt = d[a(cos t + log tan t/2)]/dt

=> dx/dt = a[d(cos t)/dt + d(log tan t/2)/dt]

=> dx/dt = a[-sin t + (1/tan t/2) * d(tan t/2)dt]

=> dx/dt = a[-sin t + cot t/2 * sec2 t/2 * d(t/2)/dt]

=> dx/dt = a[-sin t + cot t/2 * sec2 t/2 * 1/2]

=> dx/dt = a[-sin t + (cos t/2)/(sin t/2) * 1/cos2 t/2 * 1/2]

=> dx/dt = a[-sin t + 1/(2 * sin t/2 * cos t/2)]

=> dx/dt = a[-sin t + 1/sin t]

=> dx/dt = a[(-sin2 t + 1)/sin t]

=> dx/dt = a[cos2 t/sin t]

and dy/dt = d(a sin t)/dt

=> dy/dt = a * d(sin t)/dt

=> dy/dt = a cos t

Now, dy/dx = (a cos t)/[a cos2 t/sin t]

=> dy/dx = sin t /cos t

=> dy/dx = tan t

Question 9:

x = a sec θ, y = b tan θ

Answer:

The given equations are: x = a sec θ, y = b tan θ

Now, dx/dθ = d(a sec θ)/dθ

=> dx/dθ = a * d(sec θ)/dθ

=> dx/dθ = a sec θ tan θ

and dy/dθ = d(b tan θ)/dθ

=> dy/dθ = b * d(tan θ)/dθ

=> dy/dθ = b sec2 θ

Now, dy/dx = (dy/dθ)/(dx/dθ)

=> dy/dx = (b sec2 θ)/(a sec θ tan θ)

=> dy/dx = ( b/a) * sec θ cot θ

=> dy/dx = ( b/a) * (1/cos θ) * (cos θ/sin θ)

=> dy/dx = ( b/a) * (1/sin θ)

=> dy/dx = ( b/a) * cosec θ

Question 10:

x = a(cos θ + θ sin θ), y = a(sin θ - θ cos θ)

Answer:

The given equations are: x = a(cos θ + θ sin θ), y = a(sin θ - θ cos θ)

Now, dx/dθ = d{ a(cos θ + θ sin θ)}/dθ

=> dx/dθ = a[d(cos θ)/dθ + d(θ sin θ)/dθ]

=> dx/dθ = a[-sin θ + θ * d(sin θ)/dθ + sin θ * d(θ)/dθ]

=> dx/dθ = a[-sin θ + θ * cos θ + sin θ]

=> dx/dθ = aθ * cos θ

and dy/dθ = d{a(sin θ - θ cos θ)}/dθ

=> dy/dθ = a[d(sin θ)/dθ - d(θ cos θ)/dθ]

=> dy/dθ = a[cos θ + θ * d(cos θ)/dθ + cos θ * d(θ)/dθ]

=> dy/dθ = a[cos θ - θ * sin θ + cos θ]

=> dy/dθ = aθ * sin θ

Now, dy/dx = (dy/dθ)/(dx/dθ)

=> dy/dx = (aθ * sin θ)/(aθ * cos θ)

=> dy/dx = sin θ/cos θ

=> dy/dx = tan θ

Question 11:

If x = √(asin-1 t), y = √(acos-1 t), show that dy/dx = -y/x

Answer:

The given equations are: x = √(asin-1 t), y = √(acos-1 t)

=> x = (asin-1 t)1/2, y = (acos-1 t)1/2

Consider, x = (asin-1 t)1/2

Take log on both sides, we get

log x = log(asin-1 t)1/2

log x = (sin-1 t /2) log a

Differentiate w.r.t. t, we get,

Now, (1/x) * dx/dt = d[(sin-1 t /2) * log a]/dt

=> (1/x) * dx/dt = (log a /2)[d(sin-1 t)/dt]

=> (1/x) * dx/dt = (log a /2)[1/√(1 – t2)]

=> dx/dt = (x * log a /2)/√(1 – t2)

Consider, y = (acos-1 t)1/2

Take log on both sides, we get

log y = log(acos-1 t)1/2

log y = (cos-1 t /2) log a

Differentiate w.r.t. t, we get

      (1/y) * dy/dt = d[(cos-1 t /2) * log a]/dt

=> (1/y) * dy/dt = (log a /2)[d(cos-1 t)/dt]

=> (1/y) * dy/dt = (log a /2)[-1/√(1 – t2)]

=> dy/dt = (-y * log a /2)/√(1 – t2)

Now, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = [(-y * log a /2)/√(1 – t2)]/ [(x * log a /2)/√(1 – t2)]

=> dy/dx = -y/x

Hence, proved.

                                                                        Exercise 5.7

Find the second order derivatives of the functions given in Exercises 1 to 10.

Question 1:

x2 + 3x + 2

Answer:

Let y = x2 + 3x + 2

Differentiate w.r.t. t, we get

      dy/dx = d(x2 + 3x + 2)/dx

=> dy/dx = d(x2)/dx + d(3x)/dx + d(2)/dx

=> dy/dx = 2x + 3 + 0

=> dy/dx = 2x + 3

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d(2x + 3)/dx

=> d2y/dx2 = d(2x)/dx + d(3)/dx

=> d2y/dx2 = 2 + 0

=> d2y/dx2 = 2

Question 2:

x20

Answer:

Let y = x20

Differentiate w.r.t. t, we get

      dy/dx = d(x20)/dx

=> dy/dx = 20x19

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d(20x19)/dx

=> d2y/dx2 = 20 * d(x19)/dx

=> d2y/dx2 = 20 * 19 * x18

=> d2y/dx2 = 380x18

Question 3:

x * cos x

Answer:

Let y = x * cos x

Differentiate w.r.t. t, we get

      dy/dx = d(x * cos x)/dx

=> dy/dx = x * d(cos x)/dx + cos x * d(x)/dx

=> dy/dx = -x sin x + cos x

=> dy/dx = cos x – x sin x

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d(cos x – x sin x)/dx

=> d2y/dx2 = d(cos x)/dx - d(x sin x)/dx

=> d2y/dx2 = -sin x – [x * d(sin x)/dx + sin x * d(x)/dx]

=> d2y/dx2 = -sin x – [x * cos x + sin x]

=> d2y/dx2 = -sin x – x cos x - sin x

=> d2y/dx2 = -x cos x - 2sin x

=> d2y/dx2 = -(x cos x + 2sin x)

Question 4:

log x

Answer:

Let y = log x

Differentiate w.r.t. t, we get

      dy/dx = d(log x)/dx

=> dy/dx = 1/x

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d(1/x)/dx

=> d2y/dx2 = -1/x2

Question 5:

x3 * log x

Answer:

Let y = x3 * log x

Differentiate w.r.t. t, we get

      dy/dx = d(x3 * log x)/dx

=> dy/dx = x3 * d(log x)/dx + log x * d(x3)/dx

=> dy/dx = x3 * 1/x + log x * 3x2

=> dy/dx = x2 + log x * 3x2

=> dy/dx = x2(1 + 3 log x)

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d{x2(1 + 3 log x)}/dx

=> d2y/dx2 = x2 * d{(1 + 3 log x)}/dx + (1 + 3 log x) * d(x2)/dx

=> d2y/dx2 = x2 * (3/x) + (1 + 3 log x) * 2x

 => d2y/dx2 = 3x + 2x + 6x log x

=> d2y/dx2 = 5x + 6x log x

=> d2y/dx2 = x(5 + 6log x)

Question 6:

ex * sin 5x

Answer:

Let y = ex * sin 5x

Differentiate w.r.t. t, we get

      dy/dx = d(ex * sin 5x)/dx

=> dy/dx = ex * d(sin 5x)/dx + sin 5x * d(ex)/dx

=> dy/dx = ex * 5 cos 5x + sin 5x * ex

=> dy/dx = ex(sin 5x + 5 cos 5x)

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d{ex(sin 5x + 5 cos 5x)}/dx

=> d2y/dx2 = ex * d{(sin 5x + 5 cos 5x)}/dx + (sin 5x + 5 cos 5x) * d(ex)/dx

=> d2y/dx2 = ex * (5 cos 5x - 25 sin 5x) + (sin 5x + 5 cos 5x) * ex

 => d2y/dx2 = ex * (5 cos 5x - 25 sin 5x + sin 5x + 5 cos 5x)

=> d2y/dx2 = ex(10 cos 5x - 24 sin 5x)

=> d2y/dx2 = 2ex(5 cos 5x - 12 sin 5x)

Question 7:

e6x * cos 3x

Answer:

Let y = e6x * cos 3x

Differentiate w.r.t. t, we get

      dy/dx = d(e6x * cos 3x)/dx

=> dy/dx = e6x * d(cos 3x)/dx + cos 3x * d(e6x)/dx

=> dy/dx = -e6x * 3 sin 3x + 6 cos 3x * e6x

=> dy/dx = e6x(6 cos 3x - 3 sin 3x)

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d{e6x(6 cos 3x - 3 sin 3x)}/dx

=> d2y/dx2 = e6x * d{(6 cos 3x - 3 sin 3x)}/dx + (6 cos 3x - 3 sin 3x) * d(e6x)/dx

=> d2y/dx2 = e6x * (-18 sin 3x - 9 cos 3x) + (6 cos 3x - 3 sin 3x) * 6ex

=> d2y/dx2 = e6x * (-18 sin 3x - 9 cos 3x) + (36 cos 3x - 18 sin 3x) * ex

 => d2y/dx2 = e6x(-18 sin 3x - 9 cos 3x + 36 cos 3x - 18 sin 3x)

=> d2y/dx2 = e6x(27 cos 3x - 36 sin 3x)

=> d2y/dx2 = 9e6x(3 cos 3x - 4 sin 3x)

Question 8:

tan-1 x

Answer:

Let y = tan-1 x

Differentiate w.r.t. t, we get

      dy/dx = d(tan-1 x)/dx

=> dy/dx = 1/(1 + x2)

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d{1/(1 + x2)}/dx

=> d2y/dx2 = d{(1 + x2)-1}/dx

=> d2y/dx2 = (-1) * (1 + x2)-2 * d{(1 + x2)}/dx

=> d2y/dx2 = -2x/(1 + x2)2

Question 9:

log(log x)

Answer:

Let y = log(log x)

Differentiate w.r.t. t, we get

      dy/dx = d{log(log x)}/dx

=> dy/dx = (1/log x) * d(log x)/dx

=> dy/dx = (1/log x) * 1/x

=> dy/dx = 1/(x * log x)

=> dy/dx = (x * log x)-1

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d{(x * log x)-1}/dx

=> d2y/dx2 = (-1) * (x * log x)-2 * d{(x * log x)}/dx

=> d2y/dx2 = (-1) * (x * log x)-2 * [x * d(log x)/dx + log x * d(x)/dx]

=> d2y/dx2 = (-1) * (x * log x)-2 * [x * (1/x) + log x]

=> d2y/dx2 = (-1) * (x * log x)-2 * (1 + log x)

=> d2y/dx2 = -(1 + log x)/(x * log x)2

Question 10:

sin(log x)

Answer:

Let y = sin(log x)

Differentiate w.r.t. t, we get

      dy/dx = d{sin(log x)}/dx

=> dy/dx = cos(log x) * d(log x)/dx

=> dy/dx = cos(log x) * 1/x

=> dy/dx = cos(log x) /x

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d{cos(log x) /x}/dx

=> d2y/dx2 = [x * d{cos(log x)} - cos(log x) * d(x)/dx]/x2

=> d2y/dx2 = [x * {-sin(log x) * d(log x)/dx} - cos(log x) * d(x)/dx]/x2

=> d2y/dx2 = [x * {-sin(log x) * 1/x} - cos(log x)]/x2

=> d2y/dx2 = -[sin(log x) + cos(log x)]/x2

Question 11:

If y = 5 cos x – 3 sin x, prove that d2y/dx2 + y = 0

Answer:

Given, y = 5 cos x – 3 sin x

Differentiate w.r.t. t, we get

      dy/dx = d(5 cos x – 3 sin x)/dx

=> dy/dx = d(5 cos x)/dx - d(3 sin x)/dx

=> dy/dx = -5 sin x - 3 cos x

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d(-5 sin x - 3 cos x)/dx

=> d2y/dx2 = d(-5 sin x)/dx - d(3 cos x)/dx

=> d2y/dx2 = -5 cos x + 3 sin x

=> d2y/dx2 = -(5 cos x - 3 sin x)

=> d2y/dx2 = -y

=> d2y/dx2 + y = 0

Hence, proved.

Question 12:

If y = cos -1 x, find d2y/dx2 in terms of y alone.

Answer:

Given, y = cos -1 x

Differentiate w.r.t. t, we get

      dy/dx = d(cos -1 x)/dx

=> dy/dx = -1/√(1 – x2)

=> dy/dx = -(1 – x2)-1/2

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d[-(1 – x2)-1/2]/dx

=> d2y/dx2 = -(-1/2) * (1 – x2)-3/2 * d(1 – x2))/dx

=> d2y/dx2 = 1/{2√(1 – x2)}* (-2x)

=> d2y/dx2 = -x/√(1 – x2)3     …………1

Again, y = cos -1 x

=> cos y = x

Put value of x in equation 1, we get

=> d2y/dx2 = -cos y /√(1 – cos2 y)3 

=> d2y/dx2 = -cos y /√(sin2 y)3 

=> d2y/dx2 = -cos y /sin3 y

=> d2y/dx2 = -(cos y /sin y) * 1/sin2 y

=> d2y/dx2 = -cot y * cosec2 y

Question 13:

If y = 3 cos(log x) + 4 sin(log x), show that x2y2 + xy1 + y = 0

Answer:

Given, y = 3 cos(log x) + 4 sin(log x)

Differentiate w.r.t. t, we get

      dy/dx = d[3 cos(log x) + 4 sin(log x)]/dx

=> y1 = d[3 cos(log x)]/dx + d[4 sin(log x)]/dx

=> y1 = 3[-sin(log x) * d(log x)/dx] + 4[cos(log x) * d(log x)/dx]

=> y1 = 3[-sin(log x) * 1/x] + 4[cos(log x) * 1/x]

=> y1 = [4 cos(log x) – 3 sin(log x)]/x

Again, differentiate w.r.t. t, we get

=> y2 = [x * d{4 cos(log x) – 3 sin(log x)}/dx - {4 cos(log x) – 3 sin(log x)} * d(x)/dx]/x2

=> y2 = [x * d{4 cos(log x)}/dx – x * d{3 sin(log x)}/dx - {4 cos(log x) – 3 sin(log x)}]/x2

=> y2 = [x * {-4 sin(log x) * d(log x)/dx} – x * {3 cos(log x) * d(log x)/dx} - {4 cos(log x) – 3 sin(log

                x)}]/x2

=> y2 = [x * {-4 sin(log x) * 1/x} – x * {3 cos(log x) * 1/x} - {4 cos(log x) – 3 sin(log x)}]/x2

=> y2 = [-4 sin(log x) – 3 cos(log x) - {4 cos(log x) – 3 sin(log x)}]/x2

=> y2 = [-4 sin(log x) – 3 cos(log x) - 4 cos(log x) + 3 sin(log x)]/x2

=> y2 = [-sin(log x) – 7 cos(log x)]/x2

Now, x2y2 + xy1 + y

= x2 * [-sin(log x) – 7 cos(log x)]/x2 + x * [4 cos(log x) – 3 sin(log x)]/x + 3 cos(log x) + 4 sin(log x)

= -sin(log x) – 7 cos(log x)] + 4 cos(log x) – 3 sin(log x) + 3 cos(log x) + 4 sin(log x)

= 0

So, x2y2 + xy1 + y = 0

Hence, proved.

Question 14:

If y = Aemx + Benx, show that d2y/dx2 – (m + x) * dy/dx + mny = 0

Answer:

Given, y = Aemx + Benx

Differentiate w.r.t. t, we get

      dy/dx = d[Aemx + Benx]/dx

=> dy/dx = d(Aemx)/dx + d(Benx)/dx

=> dy/dx = Amemx + Bnenx

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d[Amemx + Bnenx]/dx

=> d2y/dx2 = d(Amemx)/dx + d(Bnenx)/dx

=> d2y/dx2 = Am2emx + Bn2enx

Now, d2y/dx2 – (m + n) * dy/dx + mny

= Am2emx + Bn2enx – (m + n) * (Amemx + Bnenx) + mn(Aemx + Benx)

= Am2emx + Bn2enx – (Am2emx + Bmnenx + Amnemx + Bn2enx) + Amnemx + Bmnenx

= Am2emx + Bn2enx – Am2emx - Bmnenx - Amnemx - Bn2enx + Amnemx + Bmnenx

= 0

Hence, proved.

Question 15:

If y = 500e7x + 600e-7x, show that d2y/dx2 = 49y

Answer:

Given, y = 500e7x + 600e-7x

Differentiate w.r.t. t, we get

      dy/dx = d[500e7x + 600e-7x]/dx

=> dy/dx = d(500e7x)/dx + d(600e-7x)/dx

=> dy/dx = 500 * 7 * e7x + 600 * (-7) * e-7x

=> dy/dx = 3500e7x - 4200e-7x

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d[3500e7x - 4200e-7x]/dx

=> d2y/dx2 = d(3500e7x)/dx - d(4200e-7x)/dx

=> d2y/dx2 = 3500 * 7 * e7x - 4200 * (-7) * e-7x

=> d2y/dx2 = 49(500e7x + 600e-7x )

=> d2y/dx2 = 49y

Hence, proved.

Question 16:

If ey(x + 1) = 1, show that d2y/dx2 = (dy/dx)2

Answer:

The given relationship is ey(x + 1) = 1

=> ey = 1/1(x + 1)

Taking logarithm on both the sides, we obtain

=> y = log{1/1(x + 1)}

Differentiating this relationship with respect to x, we get

      dy/dx = 1/{1/(x + 1)} * d{1/(x + 1)}/dx

=> dy/dx = (x + 1) * (-1)/(x + 1)2

=> dy/dx = -1/(x + 1)

Again, differentiate w.r.t. t, we get

      d2y/dx2 = d[-1/(x + 1)]/dx

=> d2y/dx2 = -d[1/(x + 1)]/dx

=> d2y/dx2 = -[(-1)/ (x + 1)2]

=> d2y/dx2 = 1/ (x + 1)2

=> d2y/dx2 = {-1/(x + 1)}2

=> d2y/dx2 = (dy/dx)2

Hence, proved.

Question 17:

If y = (tan-1 x)2, show that (x2 + 1)2 y2 + 2x(x2 + 1)y1 = 2

Answer:

Given, y = (tan-1 x)2

Differentiate w.r.t. t, we get

      dy/dx = d[(tan-1 x)2]/dx

=> y1 = 2 tan-1 x * d(tan-1 x)/dx

 => y1 = 2 tan-1 x * 1/(1 + x2)

=> (1 + x2)y1 = 2 tan-1 x

Again, differentiate w.r.t. t, we get

=> (1 + x2)y2 + 2xy1 = 2 * 1/(1 + x2)

=> (1 + x2)y2 + 2xy1 = 2/(1 + x2)

=> (1 + x2)2 y2 + 2x(1 + x2)y1 = 2

Hence, proved.

                                                        Exercise 5.8

Question 1:

Verify Rolle’s Theorem for the function f(x) = x2 + 2x – 8, x ∈ [-4, 2]

Answer:

The given function f(x) = x2 + 2x – 8 being a polynomial function, is continuous in [−4, 2] and is

differentiable in (−4, 2).

f(-4) = (-4)2 + 2 * (-4) – 8 = 16 – 8 – 8 = 0

f(2) = 22 + 2 * 2 – 8 = 4 + 4 – 8 = 0

So, f (−4) = f (2) = 0

=> The value of f (x) at −4 and 2 coincides.

Rolle’s Theorem states that there is a point c ∈ (−4, 2) such that f’(c) = 0

Now, f(x) = x2 + 2x – 8

f’(x) = 2x + 2

Now, f’(c) = 0

=> 2c + 2 = 0

=> c = -1, where c ∈ (−4, 2)

Hence, Rolle’s Theorem is verified for the given function.

Question 2:

Examine if Rolle’s Theorem is applicable to any of the following functions.

Can you say some thing about the converse of Rolle’s Theorem from these examples?

(i) f(x) = [x] for x ∈ [5, 9]                   (ii) f(x) = [x] for x ∈ [-2, 2]           (iii) f(x) = x2 - 1 for x ∈ [1, 2]

Answer:

By Rolle’s Theorem, for a function f : [a, b] -> R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

f (a) = f (b)

then, there exists some c ∈ (a, b) such that f’(c) = 0

Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of

the three conditions of the hypothesis.

(i) Given, f(x) = [x] for x ∈ [5, 9]

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9 ⇒ f (x) is not continuous in [5, 9].

Also, f(5) = [5] = 5

and f(9) = [9] = 9

So, f(5) ≠ f(9)

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n ∈ (5, 9).

The left hand limit of f at x = n is

limh->0- {f(n + h) – f(n)}/h = limh->0- {(n + h) – (n)}/h = limh->0- (n – 1 - n)/h = limh->0- (-1/h) = ∞

The right hand limit of f at x = n is

limh->0+ {f(n + h) – f(n)}/h = limh->0+ {(n + h) – (n)}/h = limh->0+ (n – n)/h = limh->0+ 0 = 0

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

So, f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for f(x) = [x] for x ∈ [5, 9].

(ii) f(x) = [x] for x ∈ [-2, 2]

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2

=> f (x) is not continuous in [−2, 2].

Also, f(-2) = [-2] = -2

and f(2) = [2] = 2

So, f(-2) ≠ f(2)

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n ∈ (−2, 2).

The left hand limit of f at x = n is

limh->0- {f(n + h) – f(n)}/h = limh->0- {(n + h) – (n)}/h = limh->0- (n – 1 - n)/h = limh->0- (-1/h) = ∞

The right hand limit of f at x = n is

limh->0+ {f(n + h) – f(n)}/h = limh->0+ {(n + h) – (n)}/h = limh->0+ (n – n)/h = limh->0+ 0 = 0

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

So, f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for f(x) = [x] for x ∈ [-2, 2].

(iii) f(x) = x2 - 1 for x ∈ [1, 2]

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in

(1, 2).

Also, f(1) = 12 - 1 = 1 – 1 = 0

and f(2) = 22 - 1 = 4 – 1 = 3

So, f(1) ≠ f(2)

It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for f(x) = x2 - 1 for x ∈ [1, 2].

Question 3:

If f : [-5, 5] -> R is a differentiable function and if f’(x) does not vanish anywhere, then prove that f(-5) ≠ f(5).

Answer:

It is given that f : [-5, 5] -> R is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) f is continuous on [−5, 5].

(b) f is differentiable on (−5, 5).

Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that

      f’(c) = {f(5) – f(-5)}/{5 – (-5)}

=> f’(c) = {f(5) – f(-5)}/10

=> 10f’(c) = f(5) – f(-5)

It is also given that f’(x) does not vanish anywhere.

So, f’(c) ≠ 0

=> 10 f’(c) ≠ 0

=> f(5) – f(-5) ≠ 0

=> f(5) ≠ f(-5)

Hence, proved.

Question 4:

Verify Mean Value Theorem, if f(x) = x2 – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

Answer:

The given function is f(x) = x2 – 4x – 3

f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose

derivative is 2x − 4.

f(1) = 12 – 4 * 1 – 3 = 1 – 4 – 3 = -6

f(4) = 42 – 4 * 4 – 3 = 16 – 16 – 3 = -3

So, {f(b) –f(a)}/(b - a) = {f(4) –f(1)}/(4 - 1) = {-3 – (-6)}/3 = (-3 + 6)/3 = 3/3 = 1

Mean Value Theorem states that there is a point c ∈ (1, 4) such that f’(c) = 1

=> 2c – 4 = 1

=> 2c = 5

=> c = 5/2, where c = 5/2 ∈ (1, 4)

Hence, Mean Value Theorem is verified for the given function.

Question 5:

Verify Mean Value Theorem, if f(x) = x3 – 5x2 – 3x in the interval [a, b], where a =1 b = 3. Find all c ∈ (1, 3) for which f’(c) = 0

Answer:

The given function f is f(x) = x3 – 5x2 – 3x

f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose

derivative is 3x2 − 10x − 3.

f(1) = 13 – 5 * 12 – 3 * 1 = 1 – 5 – 3 = -7

f(3) = 33 – 5 * 32 – 3 * 3 = 27 – 45 – 9 = 27 – 54 = -27

So, {f(b) –f(a)}/(b - a) = {f(3) –f(1)}/(3 - 1) = {-27 – (-7)}/2 = (-27 + 7)/2 = -20/2 = -10

Mean Value Theorem states that there exist a point c ∈ (1, 3) such that f’(c) = -10

=> 3c2 – 10c – 3 = -10

=> 3c2 – 10c – 3 + 10 = 0

=> 3c2 – 10c + 7 = 0

=> 3c2 – 3c -7c + 7 = 0

=> 3c(c - 1) – 7(c - 1) = 0

=> (c - 1)(3c - 7) = 0

=> c = 1, 7/3, where c = 7/3 ∈ (1, 3)

Hence, Mean Value Theorem is verified for the given function and c = 7/3 ∈ (1, 3) is the only

point for which f’(c) = 0

Question 6:

Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Answer:

Mean Value Theorem states that for a function f : [a, b]-> R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

then, there exists some c (a, b) such that f’(c) = {f(b) – f(a)}/(b - a)

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of

the two conditions of the hypothesis.

(i) f(x) = [x] for x ∈ [5, 9]

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

=> f (x) is not continuous in [5, 9].

The differentiability of f in (5, 9) is checked as follows:

Let n be an integer such that n ∈ (5, 9).

The left hand limit of f at x = n is

limh->0- {f(n + h) – f(n)}/h = limh->0- {(n + h) – (n)}/h = limh->0- (n – 1 - n)/h = limh->0- (-1/h) = ∞

The right hand limit of f at x = n is

limh->0+ {f(n + h) – f(n)}/h = limh->0+ {(n + h) – (n)}/h = limh->0+ (n – n)/h = limh->0+ 0 = 0

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

So, f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value

Theorem.

Hence, Mean Value Theorem is not applicable for f(x) = [x] for x ∈ [5, 9].

(ii) f(x) = [x] for x ∈ [-2, 2]

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2 ⇒ f (x) is not continuous in [−2, 2].

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n ∈ (−2, 2).

The left hand limit of f at x = n is

limh->0- {f(n + h) – f(n)}/h = limh->0- {(n + h) – (n)}/h = limh->0- (n – 1 - n)/h = limh->0- (-1/h) = ∞

The right hand limit of f at x = n is

limh->0+ {f(n + h) – f(n)}/h = limh->0+ {(n + h) – (n)}/h = limh->0+ (n – n)/h = limh->0+ 0 = 0

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

So, f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value

Theorem.

Hence, Mean Value Theorem is not applicable for f(x) = [x] for x ∈ [-2, 2].

(iii) f(x) = x2 - 1 for x ∈ [1, 2]

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in

(1, 2).

It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for f(x) = x2 - 1 for x ∈ [1, 2].

It can be proved as follows:

f(1) = 12 – 1 = 1 – 1 = 0

f(1) = 22 – 1 = 4 – 1 = 3

So, {f(b) –f(a)}/(b - a) = {f(2) –f(1)}/(2 - 1) = (3 - 0)/1 = 3

Now, f’(x) = 2x

So, f’(c) = 3

=> 2c = 3

=> c = 3/2 = 1.5 where 1.5 ∈ [1, 2]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                                                    Miscellaneous Exercise on Chapter 5

Differentiate w.r.t. x the function in Exercises 1 to 11.

Question 1:

(3x2 – 9x + 5)9

Answer:

Let y = (3x2 – 9x + 5)9

Differentiate w.r.t. x, we get

dy/dx = d(3x2 – 9x + 5)9/dx

            = 9 * (3x2 – 9x + 5)8 *d(3x2 – 9x + 5)/dx

            = 9 * (3x2 – 9x + 5)8 * (6x - 9)

            = 9 * (3x2 – 9x + 5)8 * 3 * (2x - 3)

            = 27(3x2 – 9x + 5)8(2x - 3)

Question 2:

sin3 x + cos6 x

Answer:

Let y = sin3 x + cos6 x

Differentiate w.r.t. x, we get

dy/dx = d(sin3 x + cos6 x)/dx

           = d(sin3 x)/dx + d(cos6 x)/dx

           = 3 * sin2 x * d(sin x)/dx + 6 * cos5 x * d(cos x)/dx  

           = 3 * sin2 x * cos x + 6 * cos5 x * (-sin x)    

           = 3 sin x cos x(sn x – 2 cos4 x)

 

Question 3:

(5x)3cos 2x

Answer:

Let y = (5x)3cos 2x

Taking log on both isdes, we get

log y = 3 cox 2x * log (5x)

Differentiate w.r.t. x, we get

     (1/y ) * dy/dx = d[3 cox 2x * log (5x)]/dx

=> (1/y ) * dy/dx = 3[d(cox 2x)/dx * log (5x) + cos 2x * d(log 5x)/dx]

=> (1/y ) * dy/dx = 3[-2 sin 2x * log (5x) + cos 2x * (1/5x) * d(5x)/dx]

=> (1/y ) * dy/dx = 3[-2 sin 2x * log (5x) + cos 2x * (1/5x) * 5]

=> (1/y ) * dy/dx = 3[-2 sin 2x * log (5x) + cos 2x /x]

=> (1/y ) * dy/dx = 3cos 2x /x - 6 sin 2x * log (5x)

=> dy/dx = y[3cos 2x /x - 6 sin 2x * log (5x)]

=> dy/dx = (5x)3cos 2x[3cos 2x /x - 6 sin 2x * log (5x)]

Question 4:

sin-1(x√x), 0 ≤ x ≤ 1

Answer:

Let y = sin-1(x√x)

Differentiate w.r.t. x, we get

      dy/dx = d[sin-1(x√x)]/dx

=> dy/dx = [1/√{1 – (x√x)2}] * d(x√x)/dx

=> dy/dx = [1/√(1 – x3)] * d(x3/2)/dx

=> dy/dx = [1/√(1 – x3)] * (3/2) * x3/2 - 1

=> dy/dx = [1/√(1 – x3)] * (3/2) * x1/2

=> dy/dx = 3√x/[2√(1 – x3)]

=> dy/dx = (3/2)√[x/(1 – x3)]

Question 5:

cos-1(x/2)/√(2x + 7), -2 < x < 2

Answer:

Let y = cos-1(x/2)/√(2x + 7)

Differentiate w.r.t. x by quotient rule, we get

      dy/dx = d[cos-1(x/2)/√(2x + 7)]/dx

=> dy/dx = [√(2x + 7) * d{cos-1(x/2)}/dx - cos-1(x/2) * d{√(2x + 7)}/dx]/[√(2x + 7)]2

=> dy/dx = [√(2x + 7) * {-1/√{(1 – (x/2)2} * d(x/2)/dx} – {cos-1(x/2) * 1/2√(2x + 7) * d(2x + 7)/dx}]

                    /(2x + 7)

=> dy/dx = [√(2x + 7) * -1/√(4 – x2) – cos-1(x/2) * 2/2√(2x + 7)]/(2x + 7)

=> dy/dx = -√(2x + 7)/[√(4 – x2)(2x + 7)] – cos-1(x/2)/[√(2x + 7)(2x + 7)]

=> dy/dx = -1/[√(4 – x2) √(2x + 7)] – cos-1(x/2)/(2x + 7)3/2

Question 6:

cot-1[{√(1 + sin x) + √(1 - sin x)}/{√(1 + sin x) - √(1 - sin x)}], 0 < x < π/2

Answer:

Let y = cot-1[{√(1 + sin x) + √(1 - sin x)}/{√(1 + sin x) - √(1 - sin x)}]   …………..1

Now, {√(1 + sin x) + √(1 - sin x)}/{√(1 + sin x) - √(1 - sin x)}

= {√(1 + sin x) + √(1 - sin x)2}/[{√(1 + sin x) - √(1 - sin x)} * {√(1 + sin x) + √(1 - sin x)}]

= [(1 + sin x) + (1 - sin x) + 2√{(1 + sin x) * (1 - sin x)}]/[(1 + sin x) - (1 - sin x)]

= [2 + 2√(1 – sin2 x)]/(2 sin x)

= [1 + cos x]/sin x

= [2 cos2 x/2]/(2 * sin x/2 * cos x/2]

= (cos x/2)/(sin x/2)

= cot x/2

So, equation 1 becomes

      y = cot-1 (cot x/2)

=> y = x/2

Differentiate w.r.t. x by quotient rule, we get

      dy/dx = d[x/2]/dx

=> dy/dx = 1/2

Question 7:

(log x)log x, x > 1

Answer:

Let y = (log x)log x

Taking log on both sides, we get

log y = log x * log(log x)

Differentiate w.r.t. x by quotient rule, we get

      (1/y) * dy/dx = d[log x * log(log x)]/dx

=> (1/y) * dy/dx = log(log x) * d[log x]/dx + log x * d[log(log x)]/dx

=> (1/y) * dy/dx = log(log x) * (1/x) + log x * (1/log x) * d(log x)/dx

=> (1/y) * dy/dx = log(log x) /x + log x * (1/log x) * (1/x)

=> (1/y) * dy/dx = log(log x) /x + 1/x

=> dy/dx = y[1/x + log(log x) /x]

=> dy/dx = (log x)log x[1/x + log(log x) /x]

Question 8:

cos(a cos x + b sin x), for some constant a and b.

Answer:

Let y = cos(a cos x + b sin x)

Differentiate w.r.t. x by quotient rule, we get

      dy/dx = d[cos(a cos x + b sin x)]/dx

=> dy/dx = -sin(a cos x + b sin x) * d(a cos x + b sin x)/dx

=> dy/dx = -sin(a cos x + b sin x) * [d(a cos x)/dx + d(b sin x)/dx]

=> dy/dx = -sin(a cos x + b sin x) * [-a sin x + b cos x]

=> dy/dx = sin(a cos x + b sin x).(a sin x - b cos x)

Question 9:

(sin x - cos x)(sin x - cos x), π/4 < x < 3π/4

Answer:

Let y = (sin x - cos x)(sin x - cos x)

Taking log on both sides, we get

      log y = log[(sin x - cos x)(sin x - cos x)]

=> log y = (sin x - cos x) * log(sin x - cos x)

Differentiate w.r.t. x by quotient rule, we get

      (1/y) * dy/dx = d[(sin x - cos x) * log(sin x - cos x)]/dx

=> (1/y) * dy/dx = log(sin x - cos x) * d[(sin x - cos x)]/dx + (sin x - cos x) * d[log(sin x - cos x)]/dx

 => (1/y) * dy/dx = log(sin x - cos x) * (cos x + sin x) + (sin x - cos x) * 1/(sin x - cos) * d(sin x –

                             cos x)/dx

=> dy/dx = y[log(sin x - cos x) * (cos x + sin x) + (sin x - cos x) * 1/(sin x - cos) * (cos x + sin x)]

 => dy/dx = y[log(sin x - cos x) * (cos x + sin x) + (cos x + sin x)]

=> dy/dx = y(cos x + sin x)[1 + log(sin x - cos x)]

=> dy/dx = (sin x - cos x)(sin x - cos x)(cos x + sin x)[1 + log(sin x - cos x)]

Question 10:

xx + xa + ax + aa, for some fixed a > 0 and x > 0

Answer:

Let y = xx + xa + ax + aa

Also, let xx = u, xa = v, ax = w and aa = s

So, y = u + v + w + s

=> dy/dx = du/dx + dv/dx + dw/dx + ds/dx   ………….1

Now, u = xx

=> log u = log xx

=> log u = x * log x

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = log x * d(x)/dx + x * d(log x)/dx

=> (1/u) * du/dx = log x + x * (1/x)

=> (1/u) * du/dx = log x + x

=> du/dx = u[log x + x]

=> du/dx = xx[log x + x]   …………2

      v = xa

=> dv/dx = d(xa)/dx

=> dv/dx = a * xa-1    ………..2

      w = ax

=> dw/dx = d(ax)/dx

=> dw/dx = ax * log a  ………..3

      s = aa

=> ds/dx = d(aa)/dx

=> ds/dx = 0  ………..4

From equation 1, 2, 3 and 4, we get

      dy/dx = xx[log x + x] + a * xa-1 + ax * log a + 0

 => dy/dx = xx[log x + x] + a * xa-1 + ax * log a

Question 11:

xx2- 3 + (x - 3)x2, for x > 3

Answer:

Let y = xx2- 3 + (x - 3)x2

Also, let u = xx2- 3 and v = (x - 3)x2

So, y = u + v

Differentiate w.r.t. x, we get

dy/dx = du/dx + dv/dx   ……….1

Now, u = xx2- 3

=> log u = log(xx2- 3)

=> log u = (x2 - 3) * log x

Differentiate w.r.t. x, we get

=> (1/u) * du/dx = log x * d(x2 - 3)/dx + (x2 - 3) * d(log x)/dx

 => (1/u) * du/dx = log x * 2x + (x2 - 3) * 1/x

=> (1/u) * du/dx = 2x * log x + (x2 - 3)/x

=> du/dx = u[2x * log x + (x2 - 3)/x]

=> du/dx = xx2- 3[2x * log x + (x2 - 3)/x]   ………2

Now, v = (x - 3)x2

=> log v = log(x - 3)x2

=> log v = x2 * log (x - 3)

Differentiate w.r.t. x, we get

=> (1/v) * dv/dx = log (x - 3) * d(x2)/dx + x2 * d[log (x - 3)]/dx

=> (1/v) * dv/dx = log (x - 3) * 2x + x2 * 1/(x - 3) * d (x - 3)/dx

=> (1/v) * dv/dx = 2x * log (x - 3) + x2/(x - 3)

=> dv/dx = v[2x * log (x - 3) + x2/(x - 3)]

=> dv/dx = (x - 3)x2[2x * log (x - 3) + x2/(x - 3)]     ……..3

From equation 1, 2 and 3, we get

=> dy/dx = xx2- 3[2x * log x + (x2 - 3)/x] + (x - 3)x2[2x * log (x - 3) + x2/(x - 3)]

Question 12:

Find dy/dx if y = 12(1 – cos t), x = 10(t – sin t), -π/2 < t < π/2

Answer:

Given, y = 12(1 – cos t), x = 10(t – sin t)

Now, dx/dt = d[10(t – sin t)]/dx

=> dx/dt = 10(1 – cos t)]

and dy/dt = d[12(1 – cos t)]/dx

=> dy/dt = 12 sin t

So, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = (12 sin t)/[10(1 – cos t)]

=> dy/dx = (12 * 2 sin t/2 * cos t/2)/[10 * 2 * sin2 t/2]

=> dy/dx = (6 cos t/2)/[5 sin t/2]

=> dy/dx = (6/5) cot t/2

Question 13:

Find dy/dx if y = sin-1 x + sin-1 √(1 – x2), 0 < x < 1

Answer:

Given, y = sin-1 x + sin-1 √(1 – x2)

Differentiate w.r.t. x, we get

      dy/dx = d[sin-1 x + sin-1 √(1 – x2)]/dx

=> dy/dx = d[sin-1 x]/dx + d[sin-1 √(1 – x2)]/dx

=> dy/dx = 1/√(1 – x2) + 1/√[1 – {√(1- x2)}2] * d[√(1 – x2)]/dx

=> dy/dx = 1/√(1 – x2) + 1/√[1 – (1- x2)] * 1/2√(1 – x2)  * d(1 – x2)/dx

=> dy/dx = 1/√(1 – x2) + 1/x * 1/2√(1 – x2)  * (-2x)

=> dy/dx = 1/√(1 – x2) - 1/√(1 – x2)

=> dy/dx = 0

Question 14:

If x√(1 + y) + y√(1 + x) = 0, for -1 < x < 1, prove that dy/dx = -1/(1 + x)2

 Answer:

Given, x√(1 + y) + y√(1 + x) = 0

=> x√(1 + y) = y√(1 + x)

Squaring on both sides, we get

 => x2(1 + y) = y2(1 + x)

=> x2 + x2y = y2 + xy2

=> x2 + x2y - y2 - xy2 = 0

=> x2 - y2 + x2y - xy2 = 0

=> (x - y)(x + y) + xy(x - y) = 0

=> (x - y)(x + y + xy) = 0

Since x ≠ y

=> x – y ≠ 0

So, x + y + xy = 0

=> y(1 + x) = -x

=> y = -x/(1 + x)

Differentiate w.r.t. x, we get

=> dy/dx = d[-x/(1 + x)]/dx

=> dy/dx = [(1 + x) * d(-x)/dx – (-x) * d(1 + x)/dx]/(1 + x)2

=> dy/dx = [(1 + x) * (-1) + x]/(1 + x)2

=> dy/dx = [-1 - x + x]/(1 + x)2

=> dy/dx = -1/(1 + x)2

Hence, proved.

Question 15:

If (x - a)2 + (y - b)2 = c2 for some c > 0, prove that

[1 + (dy/dx)2]3/2/(d2y/dx2) is a constant independent of a and b.

Answer:

Given, (x - a)2 + (y - b)2 = c   ……………1

Differentiate w.r.t. x, we get

=> d[(x - a)2]/dx + d[(y - b)2]/dx = d(c2)/dx

=> 2(x - a) * d(x - a)/dx + 2(y - b) * d(y - b)/dx = 0

=> 2(x - a) + 2(y - b) * dy/dx = 0

=> 2(y - b) * dy/dx = -2(x - a)

=> dy/dx = -(x - a)/(y - b)   …………2

Again differentiate w.r.t. x, we get

=> d2y/dx2 = -[(y - b) * d(x - a)/dx - (x - a) * d(y - b)/dx]/(y - b)2

=> d2y/dx2 = -[(y - b) - (x - a) * dy/dx]/(y - b)2

=> d2y/dx2 = -[(y - b) - (x - a) * {-(x - a)/(y - b)}]/(y - b)2          [From equation 1]

=> d2y/dx2 = -[(y - b)2 + (x – a)2]/(y - b)3          

=> d2y/dx2 = -c2/(y - b)3                [From equation 1]

Putting the values of dy/dx and d2y/dx2 in [1 + (dy/dx)3/2]/(d2y/dx2), we get

    [1 + {-(x - a)/(y - b)}2]3/2/[ -c2/(y - b)3]

= [1 + (x - a)2/(y - b)2]3/2/[ -c2/(y - b)3]   

= [{(y - b)2 + (x - a)2}/(y - b)2]3/2/[-c2/(y - b)3]   

= [c2/(y - b)2]3/2/[-c2/(y - b)3]                   [From equation 1]

= [c3/(y - b)3]/[-c2/(y - b)3]

= -c3/c2

= -c, which is a constant independent of a and b

Hence, proved.

Question 16:

If cos y = x cos(a + y), with cos a ≠ ±1, prove that dy/dx = cos2(a + y)/sin a

Answer:

Given, cos y = x cos(a + y)

=> x = cos y /cos(a + y)

Differentiate w.r.t. y, we get

=> dx/dy = d[cos y /cos(a + y)]/dy

=> dx/dy = [cos(a + y) * d(cos y)/dy – cos y * d{cos(a + y)}/dx]/cos2 (a + y)

=> dx/dy = [cos(a + y) * (-sin y) – cos y * {-sin(a + y)} * d(a + y)/dx]/cos2 (a + y)

=> dx/dy = [-sin y * cos(a + y) + cos y * sin(a + y)]/cos2 (a + y)

=> dx/dy = sin(a + y - y)/cos2 (a + y)

=> dx/dy = sin a /cos2 (a + y)

=> dy/dx = cos2 (a + y) /sin a

Hence, proved.

Ouestion 17:

If x = a(cos t + t * sin t) and y = a(sin t – t * cos t), find d2y/dx2

Answer:

Given, x = a(cos t + t * sin t) and y = a(sin t – t * cos t)

Differentiate w.r.t. t, we get

      dx/dt = d[a(cos t + t * sin t)]/dt

=> dx/dt = a (-sin t + t * cos t + sin t)

 => dx/dt = at cos t

And dy/dt = d[a(sin t – t * cos t)]/dt

=> dy/dt = a[cos – (t * sin t + cos t)]

 => dy/dt = at sin t

Now, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = (at sin t)/(at cos t)

=> dy/dx = tan t

Differentiate w.r.t. x, we get

=> d2y/dx2 = d(tan t)/dx

=> d2y/dx2 = sec2 t * d(t)/dx

=> d2y/dx2 = sec2 t * 1/(dx/dt)

=> d2y/dx2 = sec2 t * 1/(at * cos t)

=> d2y/dx2 = sec2 t * 1/at * sec t

=> d2y/dx2 = sec3 t /at

Question 18:

If f(x) = |x|3, show that f”(x) exists for all real x, and find it.

Answer:

|x| can be rewrite as

|x| =    x, if x ≥ 0

             -x, if x < 0

If x ≥ 0

f(x) = x3

Now, f’(x) = 3x2

And f”(x) = 6x

If x < 0

f(x) = -x3

Now, f’(x) = -3x2

And f”(x) = -6x

Hence, f”(x) exists for all real x and it can be represented as follows:

f”(x) =    6x, if x ≥ 0

              -6x, if x < 0

Question 19:

Using mathematical induction prove that d(xn)/dx = n * xn-1 for all positive integers n.

Answer:

We have to prove that d(xn)/dx = n * xn-1 for all positive integers n.

For n = 1,

P(1): d(x1)/dx = 1 * x1-1

=> d(x)/dx = x0

=> d(x)/dx = 1

So, P(n) is true for n = 1

Let P(k) is true for some positive integer k.

That is P(k): d(xk)/dx = k * xk-1

It has to be proved that P(k + 1) is also true.

Consider d(xk+1)/dx = d(x * xk)/dx

                                   = xk * d(x)/dx + x * d(xk)/dx

                                   = xk + x * k *xk-1  

                                   = xk + k *xk

                                   = (k + 1)xk

                                   = (k + 1)x(k + 1) - 1

Thus, P(k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, the statement P(n) is true for every

positive integer n.

Question 20:

Using the fact that sin(A + B) = sin A * cos B + cos A * sin B and the differentiation, obtain the sum formula for cosines.

Answer:

Given, sin(A + B) = sin A * cos B + cos A * sin B

Differentiating both sides with respect to x, we get

      d[sin(A + B)]/dx = d[sin A * cos B + cos A * sin B]/dx

=> cos(A + B) * d(A + B)/dx = d[sin A * cos B]/dx + d[cos A * sin B]/dx

=> cos(A + B) * d(A + B)/dx = d[sin A * cos B]/dx + d[cos A * sin B]/dx

=> cos(A + B) * [dA/dx + dB/dx] = cos B * d(sin A)/dx + sin A * d(cos B)/dx + sin B * d(cos A)/dx

                                                           + cos A * d(sin B)/dx

=> cos(A + B) * [dA/dx + dB/dx] = cos B * cos A * dA/dx + sin A * (-sin B) * dB/dx + sin B * (-sin

  1. A) * dA/dx + cos A * cos B * dB/dx

=> cos(A + B) * [dA/dx + dB/dx] = (cos A * cos B – sin A * sin B)[dA/dx + dB/dx]

=> cos(A + B) = cos A * cos B – sin A * sin B       

Question 21:

Does there exists a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer?

Answer:

Yes, there are some functions which are continuous everywhere but not differentiable

at exactly two points.

Let us take an example.

Let f(x) = |x - 1| + |x - 2|

Since we know that modulus functions are continuous at every point,

So there sum is also continuous at every point. But it is not differentiable at every point.

Let x = 1, 2

Now at x = 1

LHD = limx->1-  [{f(x) - f(1)}/(x - 1)]

 

        = limh->0 [{f(1 - h) - f(1)}/-h]

 

        = limh->0 [{|1 – h - 1| + |1 – h - 2| - |1 - 1|-|1 - 2|}/-h]

 

        = limh->0 [{|1 – h - 1| + |1 – h - 2| - |0|-|-1|}/-h]  

 

        = limh->0 [{|-h| + |-h - 1| - 1}/-h]

 

        = limh->0 [{h - (-h - 1) - 1}/-h] 

 

        = limh->0 [{h + h + 1 - 1}/-h]

 

        = limh->0 [{2h}/-h]

 

        = -2

RHD = limx->1+  [{f(x) - f(1)}/(x - 1)] 

 

         = limh->0 [{f(1 + h) - f(1)}/h]

 

        = limh->0 [{|1 + h - 1| + |1 + h - 2| - |1 - 1|-|1 - 2|}/h]

 

        = limh->0 [{|1 + h - 1| + |1 + h - 2| - |0|-|-1|}/h]  

 

        = limh->0 [{|h| + |h - 1| - 1}/h]

 

        = limh->0 [{h - (h - 1) - 1}/h] 

        = limh->0 [{h - h + 1 - 1}/h]

 

        = limh->0 [0/h]

 

        = 0

Since LHD ≠ RHD

So given function is not diffenetiable at x = 1.

Similarly, we can show that the given function is not differentiable at x = 2.

Question 22:                                                                                                                       

If            f(x)    g(x)      h(x)                                          f’(x)      g’(x)    h’(x)                 

     y =       l         m        n      , prove that dy/dx =   l             m         n              

                 a         b         c                                            a             b          c              

Answer:

Given,

               f(x)    g(x)      h(x)                                             

     y =       l         m        n                             

                 a         b         c                    

=> y = (mc - nb)f(x) – (lc - na)g(x) - (lb - ma)h(x)

Differentiating both sides with respect to x, we get

      dy/dx = d[(mc - nb)f(x) – (lc - na)g(x) - (lb - ma)h(x)]/dx

=> dy/dx = d[(mc - nb)f(x)]/dx – d[(lc - na)g(x)]/dx – d[(lb - ma)h(x)]/dx

=> dy/dx = (mc - nb)f’(x) – (lc - na)g’(x) – (lb - ma)h’(x)

=> dy/dx =   f’(x)    g’(x)      h’(x)                                             

                        l          m           n                             

                        a          b           c                  

Question 23:

If y = eacos-1 x,-1 ≤ x ≤ 1, show that (1 – x2)d2y/dx2 – x * dy/dx – a2y = 0

Answer:

It is given that y = eacos-1 x

Taking log on both sides, we get

=> log y = log(eacos-1 x)

=> log y = a cos-1 x * log e

=> log y = a cos-1 x

Differentiating both sides with respect to x, we get

      dy/dx = d[a cos-1 x]/dx

=> (1/y) * dy/dx = a * (-1)/√(1 – x2)

=> dy/dx = -ay/√(1 – x2)

Squaring on both sides, we get

=> (dy/dx)2 = a2y2/(1 – x2)

=> (1 – x2)(dy/dx)2 = a2y2

Again, differentiating both sides with respect to x, we get

=> d[(1 – x2)(dy/dx)2]/dx = d(a2y2)/dx

=> (dy/dx)2 * d[(1 – x2)]/dx] + (1 – x2) * d[(dy/dx)2]/dx = d(a2y2)/dx

=> (dy/dx)2 * (-2x) + (1 – x2) * 2 * dy/dx * d(dy/dx)/dx = 2a2y * dy/dx

=> (dy/dx)2 * (-2x) + (1 – x2) * 2 * dy/dx * d2y/dx2 = 2a2y * dy/dx

=> -x * (dy/dx) + (1 – x2) * dy/dx = a2y         [Since dy/dx ≠ 0]

=> (1 – x2)d2y/dx2 – x * dy/dx – a2y = 0

Hence, proved.

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