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

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.

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|

(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.

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?

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

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 }

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 }

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 }

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 }

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 }

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 }

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?

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 }

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}

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 }

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.

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?

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.

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?

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

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,

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

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?

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 }

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

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

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

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

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.

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.

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.

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.

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|.

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)

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)

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)

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)

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)

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

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)

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

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.

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.

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

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

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

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

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

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

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

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

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)}

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

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

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)}

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)}

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

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

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

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

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

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)

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)

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

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

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

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

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

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

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)}]

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

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

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

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)

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

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)

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

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)

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

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

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

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

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)

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).

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?

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.

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

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

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

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

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

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)

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)

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

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

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)

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

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

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

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

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

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

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

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

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

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)

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)

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

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.

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

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

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

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

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

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]

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]

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).

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.

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

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.

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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.

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

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

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.

|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.

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.

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?

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

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

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.

.