Class 11 - Maths - Limits Derivatives

Exercise 13.1

Evaluate the following limits in Exercises 1 to 22.

Question 1:

limx->3 (x + 3)

Given, limx->3 (x + 3) = 3 + 3 = 6

Question 2:

limx-> π (x – 22/7)

Given, limx->π (x – 22/7) = π – 22/7

Question 3:

limr-> 1 (πr2)

Given, limr-> 1 (πr2) = π(1)2 = π

Question 4:

limx-> 4 (4x + 3)/(x - 2)

Given, limx-> 4 (4x + 3)/(x - 2) = (4 * 4 + 3)/(4 - 2)

= (16 + 3)/2

= 19/2

Question 5:

limx-> -1 (x10 + x5 + 1)/(x - 1)

Given, limx-> -1 (x10 + x5 + 1)/(x - 1) = {(-1)10 + (-1)5 + 1)}/(-1 - 1)

= (1 – 1 + 1)/(-2)

= -1/2

Question 6:

limx-> 0 {(x + 1)5 - 1)}/x

Given, limx-> 0 {(x + 1)5 - 1)}/x

Put x + 1 = y so that y → 1 as x → 0.

Now, limx-> 0 {(x + 1)5 - 1)}/x = limy->1 (y5 - 1))/(y - 1)

= limy->1 (y5 - 15))/(y - 1)

= 5 * 15-1                            [Since limx->a (xn – an))/(x - a) = n * an-1

= 5

So, limx-> 0 {(x + 1)5 - 1)}/x = 5

Question 7:

limx-> 2 {(3x2 – x - 10)/(x2 - 4)}

At x = 2, the value of the given rational function takes the form 0/0

Now, limx-> 2 {(3x2 – x - 10)/(x2 - 4)} = limx-> 2 [{(x - 2)(3x + 5)}/{(x - 2)(x + 2)}]

= limx-> 2 [(3x + 5)/(x + 2)]

= (3 * 2 + 5)/(2 + 2)

= 11/4

Question 8:

limx-> 3 {(x4 – 81)/(2x2 – 5x - 3)}

At x = 2, the value of the given rational function takes the form 0/0

Now, limx-> 3 {(x4 – 81)/(2x2 – 5x - 3)} = limx-> 3 [{(x - 3)(x + 3)(x2 + 9)}/{(x - 3)(2x + 1)}]

= limx-> 3 [{(x + 3)(x2 + 9)}/(2x + 1)]

= {(3 + 3)(32 + 9)}/(2 * 3 + 1)

= (6 * 18)/7

= 108/7

Question 9:

limx-> 0 {(ax + b)/(cx + 1)}

Given limx-> 0 {(ax + b)/(cx + 1)} = (a * 0 + b)/(c * 0 + 1) = b

Question 10:

limz-> 1 {(z1/3 - 1)/(z1/6 - 1)}

Given, limz-> 1 {(z1/3 - 1)/(z1/6 - 1)}

At z = 1, the value of the given function takes the form 0/0

Put z1/6 = x so that z -> 1 as x -> 1

Now, limz-> 1 {(z1/3 - 1)/(z1/6 - 1)} = limx-> 1 {(x2 - 1)/(x - 1)}

= limx-> 1 {(x2 - 12)/(x - 1)}

= 2 * 12-1                            [Since limx->a (xn – an))/(x - a) = n * an-1]

= 2

So, limz-> 1 {(z1/3 - 1)/(z1/6 - 1)} = 2

Question 11:

limx-> 1 {(ax2 + bx + c)/(cx2 + bx + a)}, a + b + c ≠ 0

Given limx-> 1 {(ax2 + bx + c)/(cx2 + bx + a)} = (a * 1 + b * 1 + c)/(c * 1 + b * 1 + a)

= (a + b + c)/(a + b + c)

= 1

Question 12:

limx-> -2 {(1/x + 1/2)/(x + 2)}

Given, limx-> -2 {(1/x + 1/2)/(x + 2)}

At x = -2, the value of the given function takes the form 0/0

Now, limx-> -2 {(1/x + 1/2)/(x + 2)} = limx-> -2 [{(2 + x)/(2x)}/(x + 2)]

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

= 1/{2 * (-2)}

= -1/4

Question 13:

limx-> -0 (sin ax /bx)

Given, limx-> -0 (sin ax /bx)

At x = 0, the value of the given function takes the form 0/0

Now, limx-> -0 (sin ax /bx) = limx-> -0 {(sin ax /ax) * (ax/bx)}

= limx-> -0 {(sin ax /ax) * (a/b)}

= (a/b) * limx-> -0 (sin ax /ax)

= (a/b) * 1             [Since limy-> -0 (sin y /y) = 1]

= a/b

Question 14:

limx-> -0 (sin ax /sin bx), a, b ≠ 0

Given, limx-> -0 (sin ax /sin bx)

At x = 0, the value of the given function takes the form 0/0

Now, limx-> -0 (sin ax /sin bx) = limx-> -0 [{(sin ax /ax)/(sin bx/bx)} * (ax/bx)]

= (a/b) * [{limx-> -0 (sin ax /ax)}/{limx-> -0 (sin ax /ax)}]

= (a/b) * 1             [Since limy-> -0 (sin y /y) = 1]

= (a/b)

Question 15:

limx-> π [sin (π - x) /{π(π - x)}]

Given, limx-> π [sin (π - x)/{π(π - x)}]

It is seen that x -> π => π – x = 0

Now, limx-> π [sin (π - x)/{π(π - x)}] = (1/ π) * lim(π-x)->0 [sin (π - x) /(π - x)]

= (1/ π) * 1              [Since limy-> -0 (sin y /y) = 1]

= 1/ π

Question 16:

limx-> 0 [cos x /(π - x)]

Given, limx-> 0 [cos x/(π - x)] = [cos 0/(π - 0)] = 1/π

Question 17:

limx-> 0 [(cos 2x – 1)/ (cos x – 1)]

Given, limx-> 0 [(cos 2x – 1)/ (cos x – 1)]

At x = 0, the value of the given function takes the form 0/0

Now, limx-> 0 [(cos 2x – 1)/ (cos x – 1)] = limx-> 0 [(1 - 2sin2 x – 1)/ (1 - 2sin2 x/2 – 1)]

= limx-> 0 [sin2 x /sin2 x/2]

= limx-> 0 [{(sin2 x /x2)/(sin2 x/2 /(x/2)2} * x2/(x2/4)]

= 4 * limx-> 0 [(sin x /x)2/{(sin x/2 /(x/2)}2]

= 4 * limx-> 0 [(sin x /x)2]/[limx-> 0 [{(sin x/2 /(x/2)}2]

= 4 * (12/12)                    [Since limy-> -0 (sin y /y) = 1]

= 4

Question 18:

limx-> 0 [(ax + x * cos x)/(b * sin x)]

Given, limx-> 0 [(ax + x * cos x)/(b * sin x)]

At x = 0, the value of the given function takes the form 0/0

Now, limx-> 0 [(ax + x * cos x)/(b * sin x)] = (1/b) * limx-> 0 [x(a + cos x)/sin x]

= (1/b) * limx-> 0 [x/sin x] * limx-> 0 [a + cos x]

= (1/b) * limx-> 0 [1/(sin x /x)] * [a + cos 0]

= (1/b) * 1 * [a + 1]            [Since limy-> -0 (sin y /y) = 1]

= (a + 1)/b

Question 19:

limx-> 0 [x * sec x]

Given, limx-> 0 [x * sec x] = limx-> 0 [x/cos x] = 0/ cos 0 = 0/1 = 0

Question 20:

limx-> 0 [(sin ax + bx)/(ax + sin bx)], a, b, a + b ≠ 0

Given, limx-> 0 [(sin ax + bx)/(ax + sin bx)]

At x = 0, the value of the given function takes the form 0/0

Now, limx-> 0 [(sin ax + bx)/(ax + sin bx)]

= limx-> 0 [{(sin ax/ax) * ax + bx)}/{ax + bx * sin bx /bx}]

= {limx-> 0 (sin ax/ax) * limx-> 0 ax + limx-> 0 bx}/{limx-> 0 ax + limx-> 0 bx * limx-> 0 sin bx /bx}

= {limx-> 0 (sin ax/ax) * a * limx-> 0 x + b * limx-> 0 x}/{a * limx-> 0 x + b * limx-> 0 x * limx-> 0 sin bx /bx}

= (1 + a * 0 + b * 0)/(a * 0 + b * 0 + 1)                          [Since limy-> -0 (sin y /y) = 1]

= 1

Question 21:

limx-> 0 (cosec x – cot x)

Given, limx-> 0 (cosec x – cot x)

At x = 0, the value of the given function takes the form ∞ - ∞

Now, limx-> 0 (cosec x – cot x) = limx-> 0 (1/sin x – cos x / sin x)

= limx-> 0 {(1 – cos x)/sin x}

= limx-> 0 [{(1 – cos x)/x}/(sin x/x)]

= limx-> 0 {(1 – cos x)/x}/{limx-> 0 (sin x/x)}

= 0/1              [Since limx-> 0 (1 – cos x)/x = 0 and limx-> 0 (sin x/x) = 1]

= 0

Question 22:

limx-> π/2 [tan 2x/(x – π/2)]

Given, limx-> π/2 [tan 2x/(x – π/2)]

At x = π/2, the value of the given function takes the form 0/0

Put x – π/2 = y so that x -> π/2, y -> 0

Now, limx-> π/2 [tan 2x/(x – π/2)] = limy-> 0 [tan 2(y + π/2) /y]

= limy-> 0 [tan (2y + π) /y]

= limy-> 0 [tan 2y /y]                  [Since tan (π + 2y) = tan 2y]

= limy-> 0 [sin 2y /(y * cos 2y)]

= limy-> 0 [(sin 2y /2y) * (2/cos 2y)]

= limy-> 0 (sin 2y /2y) * limy-> 0 (2/cos 2y)

= 1 * (2/cos 0)                   [Since limx-> -0 (sin x /x) = 1]

= 1 * (2/1)

= 2

Question 23:

Find limx-> 0 f(x) and limx-> 1 f(x) where f(x) =     2x + 3, x ≤ 0

3(x + 1), x > 0

The given function is   f(x) =     2x + 3, x ≤ 0

3(x + 1), x > 0

Now, lim x -> 0- f(x) = lim x -> 0 (2x + 3) = 2 * 0 + 3 = 3

lim x -> 0+ f(x) = lim x -> 0 3(x + 1) = 3(0 + 1) = 3

So, lim x -> 0- f(x) = lim x -> 0+ f(x) = lim x -> 0 f(x) = 3

lim x -> 1- f(x) = lim x -> 1 3(x + 1) = 3(1 + 1) = 3 * 2 = 6

lim x -> 1+ f(x) = lim x ->1 3(x + 1) = 3(1 + 1) = 3 * 2 = 6

So, lim x -> 1- f(x) = lim x -> 1+ f(x) = lim x -> 1 f(x) = 6

Question 24:

Find limx-> 1 f(x) where f(x) =     x2 - 1, x ≤ 1

-x2 - 1, x > 1

Given function is f(x) =    x2 - 1, x ≤ 1

-x2 - 1, x > 1

lim x -> 1- f(x) = lim x -> 1 (x2 – 1) = 12 - 1 = 1 – 1 = 0

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

Now, lim x -> 1- f(x) ≠ lim x -> 1+ f(x)

Hence, lim x -> 1 f(x) does not exist.

Question 25:

Find limx-> 0 f(x) where f(x) =     |x|/x,   x ≠ 0

0,         x = 0

Given function is f(x) =   |x|/x,  x ≠ 0

0,        x = 0

Now, lim x -> 0- f(x) = lim x -> 0- [|x|/x]

= lim x -> 0 (-x/x)                   [When x is negative, |x| = -x]

= lim x -> 0 (-1)

= -1

Again, lim x -> 0+ f(x) = lim x -> 0+ [|x|/x]

= lim x -> 0 (x/x)                   [When x is negative, |x| = x]

= lim x -> 0 (1)

= 1

Since lim x -> 0- f(x) ≠ lim x -> 0+ f(x)

Hence, lim x -> 0 f(x) does not exist.

Question 26:

Find limx-> 0 f(x) where f(x) =      x/|x|, x ≠ 0

0, x = 0

Given function is f(x) =   x/|x|, x ≠ 0

0, x = 0

Now, lim x -> 0- f(x) = lim x -> 0- [x/|x|]

= lim x -> 0 (-x/x)                   [When x is negative, |x| = -x]

= lim x -> 0 (-1)

= -1

Again, lim x -> 0+ f(x) = lim x -> 0+ [x/|x|]

= lim x -> 0 (x/x)                   [When x is negative, |x| = x]

= lim x -> 0 (1)

= 1

Since lim x -> 0- f(x) ≠ lim x -> 0+ f(x)

Hence, lim x -> 0 f(x) does not exist.

Question 27:

Find limx-> 5 f(x) where f(x) = |x| - 5

Given function is f(x) = |x| - 5

Now, lim x ->5- f(x) = lim x -> 5- [|x| - 5]

= lim x -> 0 (x - 5)                   [When x > 0, |x| = x]

= lim x -> 0 (5 - 5)

= 0

Again, lim x -> 5+ f(x) = lim x -> 5+ [|x| - 5]

= lim x -> 5 (x - 5)                   [When x > 0, |x| = x]

= lim x -> 0 (5 - 5)

= 0

Since lim x ->5- f(x) = lim x ->5+ f(x) = 0

Hence, lim x -> 5 f(x) = 0

Question 28:

Suppose                       a + bx, if x < 1

f(x) =        4,         if x = 0

b – ax, if x > 1

and limx->1 f(x) = f(1). What are the possible values of a and b?

Given function is:

a + bx, if x < 1

f(x) =        4,         if x = 0

b – ax, if x > 1

Now, lim x ->1- f(x) = lim x ->1 (a + bx) = a + b

and lim x ->1+ f(x) = lim x ->1 (b - ax) = b – a

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

=> lim x ->1- f(x) = lim x ->1+ f(x) = lim x ->1 f(x) = f(1)

=> a + b = 4 and b – a = 4

On solving these two equations, we get

a = 0, b = 4

Hence, the possible values of a and b are 0 and 4 respectively.

Question 29:

Let a1, a2, a3, ………………, an be fixed real numbers and define a function

f(x) = (x – a1) (x – a2) (x – a3)…………… (x – an)

What is limx – a1 f(x)? For some a ≠ a1, a2, a3, ………………, an, compute limx – a f(x)

Given function is:

f(x) = (x – a1) (x – a2) (x – a3)…………… (x – an)

Now, limx –> a1 f(x) = limx –> a1 [(x – a1)(x – a2) (x – a3)…………… (x – an)]

= {limx –> a1 (x – a1)}{ limx –> a1 (x – a2)}………{ limx –> a1 (x – an)]

= (a1 – a1)(a1 – a2)(a1 – a3)…………… (a1 – an)

= 0

So, limx –> a1 f(x) = 0

Again, limx –> a f(x) = limx –> a [(x – a1) (x – a2) (x – a3)…………… (x – an)]

= {limx –> a (x – a1)}{limx –> a (x – a2)}………{limx –> a (x – an)]

= (a – a1) (a – a2) (a – a3)…………… (a – an)

So, limx –> a f(x) = (a – a1) (a – a2) (a – a3)…………… (a – an)

Question 30:

|x| + 1, if x < 0

If f(x) =        0,          if x = 0

|x| - 1, if x > 0

For what value(s) of a does limx->a f(x) exists?

Given function is

|x| + 1, if x < 0

f(x) =        0,          if x = 0

|x| - 1, if x > 0

When a = 0,

lim x ->0- f(x) = lim x -> 0- [|x| + 1]

= lim x -> 0 (-x + 1)                   [When x < 0, |x| = -x]

= 1

lim x ->0+ f(x) = lim x -> 0+ [|x| - 1]

= lim x -> 0 (x - 1)                   [When x > 0, |x| = x]

= -1

Since lim x ->0- f(x) ≠ lim x ->0+ f(x) = 0

Hence, lim x -> 0 f(x) does not exists.

When a < 0

lim x ->a- f(x) = lim x -> a- [|x| + 1]

= lim x -> a (-x + 1)                   [When x < a < 0, |x| = -x]

= -a + 1

lim x ->a+ f(x) = lim x -> a+ [|x| + 1]

= lim x -> a (-x + 1)                   [When a < x < 0, |x| = -x]

= -a + 1

Since lim x ->a- f(x) = lim x ->a+ f(x) = -a + 1

Thus, lim x -> a f(x) exists at x = a where a < 0

When a > 0

lim x ->a- f(x) = lim x -> a- [|x| - 1]

= lim x -> a (x - 1)                   [When 0 < x < a, |x| = x]

= a - 1

lim x ->a+ f(x) = lim x -> a+ [|x| - 1]

= lim x -> a (x - 1)                   [When 0 < a < x, |x| = x]

= a - 1

Since lim x ->a- f(x) = lim x ->a+ f(x) = a - 1

Thus, lim x -> a f(x) exists at x = 0 where a > 0

Question 31:

If the function f(x) satisfies lim x ->1 [f(x) – 2]/[x2 - 1] = π, evaluate lim x ->1 f(x).

Given, lim x ->1 [f(x) – 2]/[x2 - 1] = π

=> lim x ->1 [f(x) – 2]/ lim x ->1 [x2 - 1] = π

=> lim x ->1 [f(x) – 2] = π * lim x ->1 [x2 - 1]

=> lim x ->1 [f(x) – 2] = π * [12 - 1]

=> lim x ->1 [f(x) – 2] = 0

=> lim x ->1 f(x) – lim x ->1 2 = 0

=> lim x ->1 f(x) – 2 = 0

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

Hence, lim x ->1 f(x) = 2

Question 32:

mx2 + n, if x < 0

If f(x) =       nx + m,   if 0 ≤ x ≤ 1

nx3 + m, if x > 1

For what integers m and n does limx->0 f(x) and limx->1 f(x) exists?

Given function is:

mx2 + n, if x < 0

If f(x) =       nx + m,   if 0 ≤ x ≤ 1

nx3 + m, if x > 1

lim x ->0- f(x) = lim x -> 0 (mx2 + n)

= (m * 0 + n)

= n

lim x ->0+ f(x) = lim x -> 0+ (nx + m)

= (n * 0 + m)

= m

Thus, lim x ->0 f(x) exists if m = n

lim x ->1- f(x) = lim x -> 1 (nx + m)

= (n * 1 + m)

= m + n

lim x ->1+ f(x) = lim x -> 1 (nx3 + m)

= (n * 1 + m)

= m + n

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

Thus, lim x ->1 f(x) exists for any integral value of m and n.

Exercise 13.2

Question 1:

Find the derivative of x2 – 2 at x = 10.

Let f(x) = x2 – 2

Now, df(x)/dx = d(x2 – 2)/dx

=> f’(x) = 2x

Now, at x = 10

=> f’(10) = 2 * 10 = 20

Thus, the derivative of x2 – 2 at x = 10 is 20.

Question 2:

Find the derivative of 99x at x = 100.

Let f(x) = 99x

Now, df(x)/dx = d(99x)/dx

=> f’(x) = 99

Now, at x = 100

=> f’(100) = 99

Thus, the derivative of 99x at x = 100 is 99.

Question 3:

Find the derivative of x at x = 1.

Let f(x) = x

Now, df(x)/dx = d(x)/dx

=> f’(x) = 1

Now, at x = 1

=> f’(1) = 1

Thus, the derivative of x at x = 1 is 1.

Question 4:

Find the derivative of the following functions from first principle.

(i) x3 – 27                    (ii) (x – 1)(x – 2)                 (iii) 1/x2                              (iv) (x + 1)/(x - 1)

(i) Let f(x) = x3 – 27

Now, from the first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{(x + h)3 – 27} – (x3 – 27)}/h]

= limh->0 [{x3 + h3 + 3x2 h + 3xh2 – 27 – x3 + 27}/h]

= limh->0 [{h3 + 3x2 h + 3xh2}/h]

= limh->0 [h2 + 3x2 + 3xh]

= 0 + 3x2 + 3x * 0

= 3x2

(ii) Let f(x) = (x – 1)(x – 2)

Now, from the first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{(x + h – 1)(x + h – 2) – (x – 1)(x – 2)}/h]

= limh->0 [{(x2 + hx – 2x + hx + h2 – 2h – x – h + 2) - (x2 - 2x – x + 2)}/h]

= limh->0 [{2hx + h2 – 3h}/h]

= limh->0 [2x + h - 3]

= 2x + 0 - 3

= 2x – 3

(iii) Let f(x) = 1/x2

Now, from the first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{1/(x + h)2 – 1/x2)}/h]

= limh->0 [{x2 - (x + h)2}/{h(x + h)2 x2}]

= limh->0 [{x2 - x2 - h2 – 2hx)}/{h(x + h)2 x2}]

= limh->0 [{-h2 – 2hx)}/{h(x + h)2 x2}]

= limh->0 [{-h – 2x)}/{(x + h)2 x2}]

= limh->0 [{0 – 2x)}/{(x + 0)2 x2}]

= -2x/x4

= -2/x3

(iv) Let f(x) = (x + 1)/(x - 1)

Now, from the first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{(x + h + 1)/(x + h - 1) – (x + 1)/(x – 1)}/h]

= limh->0 [{(x - 1)(x + h + 1) – (x + 1)(x + h - 1)}/{h(x - 1)/(x + h – 1)}]

= limh->0 [{(x2 + hx + x – x - h - 1) – (x2 + hx - x + x + h - 1)}/{h(x - 1)/(x + h – 1)}]

= limh->0 [(-2h)/{h(x - 1)/(x + h – 1)}]

= limh->0 [(-2)/{(x - 1)/(x + h – 1)}]

= (-2)/{(x - 1)/(x + 0 – 1)}

= -2/{(x - 1)(x - 1)}

= -2/(x - 1)2

Question 5:

For the function f(x) = x100/100 + x99/99 + …….. + x2/2 + x + 1

Prove that f’(1) = 100f’(0)

Given, f(x) = x100/100 + x99/99 + …….. + x2/2 + x + 1

df(x)/dx = d[x100/100 + x99/99 + …….. + x2/2 + x + 1]/dx

=> f’(x) = d[x100/100]/dx + d[x99/99]/dx + …….. + d[x2/2]/dx + d[x]/dx + d[1]/dx

=> f’(x) = 100x99/100 + 99x98/99 + …….. + 2x/2 + 1 + 0

=> f’(x) = x99 + x98 + …….. + x + 1

Now, at x = 0, f’(0) = 1

At x = 1

=> f’(1) = 199 + 198 + …….. + 1 + 1

= [1 + 1 + 1 + ………100 times]

= 1 * 100

= 100

Hence, f’(1) = 100f’(0)

Question 6:

Find the derivative of xn + axn-1 + a2 xn-2 + ………+ an-1 x + an for some fixed real number a.

Let f(x) = xn + axn-1 + a2 xn-2 + ………+ an-1 x + an

Now, df(x)/dx = d[xn + axn-1 + a2 xn-2 + ………+ an-1 x + an]/dx

=> f’(x) = d[xn ]/dx + d[axn-1 ]/dx + d[a2 xn-2]/dx + ………+ d[an-1 x]/dx + d[an]/dx

=> f(x) = nxn-1 + a(n - 1)xn-2 + a2(n - 2)xn-3 + ………+ an-1 + an (0)

=> f(x) = nxn-1 + a(n - 1)xn-2 + a2(n - 2)xn-3 + ………+ an-1

Question 7:

For some constants a and b, find the derivative of

(i) (x – a)(x – b)                (ii) (ax2 + b)2                          (iii) (x - a)/(x - b)

(i) Let f(x) = (x – a)(x – b)

=> f(x) = x2 – ax – bx + ab

Now, df(x)/dx = d(x2 – ax – bx + ab)/dx

=> f’(x) = d(x2)/dx– d(ax) /dx – d(bx) /dx + d(ab)/dx

=> f’(x) = 2x – a – b + 0

=> f’(x) = 2x – (a + b)

(ii) Let f(x) = (ax2 + b)2

=> f(x) = a2 x4 + b2 + 2abx2

Now, df(x)/dx = d(a2 x4 + b2 + 2abx2)/dx

=> f’(x) = d(a2 x4)/dx + d(b2)/dx + d(2abx2)/dx

=> f’(x) = 4a2 x3 + 0 + 2ab * 2x

=> f’(x) = 4a2 x3 + 4abx

=> f’(x) = 4ax(ax2 + b)

(iii) Let f(x) = (x - a)/(x - b)

Now, df(x)/dx =d{(x - a)/(x - b)}/dx

=> f’(x) = {(x - b) * d(x - a)/dx - (x - a) * d(x - b)/dx}/(x - b)2            [By quotient rule]

=> f’(x) = {(x - b) * 1 - (x - a) * 1}/(x - b)2

=> f’(x) = (x - b - x + a)/(x - b)2

=> f’(x) = (a – b)/(x - b)2

Question 8:

Find the derivative of (xn - an)/(x - a) for some constant a.

Let f(x) = (xn - an)/(x - a)

Now, df(x)/dx = d{(xn - an)/(x - a)}/dx

=> f’(x) = {(x - a) * d(xn - an)/dx - (xn - an) * d(x - a)/dx}/(x - a)2            [By quotient rule]

=> f’(x) = {(x - a) * (nxn-1 - 0) - (xn - an) * 1}/(x - a)2

=> f’(x) = {(x - a) * nxn-1 - (xn - an)}/(x - a)2

=> f’(x) = (nxn - anxn-1 - xn + an)/(x - a)2

Question 9:

Find the derivative of

(i) 2x – 3/4                           (ii) (5x3 + 3x – 1)(x – 1)                        (iii) x–3(5 + 3x)

(iv) x5(3 – 6x–9)                    (v) x–4 (3 – 4x–5)                                     (vi) 2/(x + 1) - x2/(3x - 1)

(i) Let y = 2x – 3/4

Now, dy/dx = d(2x – 3/4)/dx

=> dy/dx = d(2x)/dx  - d(3/4)/dx

=> dy/dx = 2 – 0

=> dy/dx = 2

(ii) Let y = (5x3 + 3x – 1)(x – 1)

Now, dy/dx = d{(5x3 + 3x – 1)(x – 1)}/dx

=> dy/dx = (5x3 + 3x – 1) * d(x – 1)/dx + (x - 1) * d(5x3 + 3x – 1)/dx

=> dy/dx = (5x3 + 3x – 1) * 1 + (x - 1) * (3 * 5x2 + 3)

=> dy/dx = (5x3 + 3x – 1) + (x - 1)(15x2 + 3)

=> dy/dx = 5x3 + 3x – 1 + 15x3 + 3x - 15x2 – 3

=> dy/dx = 20x3 - 15x2 + 6x – 4

(iii) Let y = x–3(5 + 3x)

=> y = 5x–3 + 3x–3+1

=> y = 5x–3 + 3x–2

Now dy/dx = d(5x–3 + 3x–2)/dx

=> dy/dx = d(5x–3)/dx + d(3x–2)/dx

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

=> dy/dx = -15x–4 - 6x–3

=> dy/dx = -15/x4 – 6/x3

=> dy/dx = -3(5 + 2x)/x4

(iv) Let y = x5(3 – 6x–9)

=> y = 3x5 – 6x–9 + 5

=> y = 3x5 – 6x–4

Now, dy/dx = d(3x5 – 6x–4)/dx

=> dy/dx = d(3x5)/dx – d(6x–4)/dx

=> dy/dx = 5 * 3x5-1 – (-4) * 6x–4-1

=> dy/dx = 15x4 + 24x–5

=> dy/dx = 15x4 + 24/x5

(v) Let y = x–4 (3 – 4x–5)

=> y = 3x–4 – 4x–5-4

=> y = 3x–4 – 4x–9

Now, dy/dx = d(3x–4 – 4x–9)/dx

=> dy/dx = d(3x–4)/dx – d(4x–9)/dx

=> dy/dx = (-4) * 3x–4-1 – (-9) * 4x–9-1

=> dy/dx = -12x–5 + 36x–10

=> dy/dx = -12/x5 + 36/x10

(vi) Let y = 2/(x + 1) - x2/(3x - 1)

Now, dy/dx = d{2/(x + 1) - x2/(3x - 1)}/dx

=> dy/dx = d{2/(x + 1)}/dx – d{x2/(3x - 1)}/dx

=> dy/dx = [{(x + 1) * d(2)/dx – 2 * d(x + 1)/dx}/(x + 1)2]

- [{(3x - 1) * d(x2)/dx – x2 * d(3x - 1)/dx}/(3x - 1)2]

=> dy/dx = [{(x + 1) * 0 – 2 * 1}/(x + 1)2] - [{(3x - 1) * 2x – 3x2}/(3x - 1)2]

=> dy/dx = -2/(x + 1)2 - [6x2 – 2x - 3x2}/(3x - 1)2]

=> dy/dx = -2/(x + 1)2 – (3x2 – 2x)/(3x - 1)2

=> dy/dx = -2/(x + 1)2 – x(3x – 2)/(3x - 1)2

Question 10:

Find the derivative of cos x from first principle.

Let f(x) = cos x

Now, from the first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{cos(x + h) – cos x}/h]

= limh->0 [{cos x * cos h – sin x * sin h – cos x}/h]

= limh->0 [{cos x (cos h – 1) – sin x * sin h}/h]

= limh->0 [{cos x (cos h – 1)}/h] – limh->0 [(sin x * sin h)/h]

= cos x * limh->0 [(cos h – 1)/h] – sin x * limh->0 [(sin h)/h]

= cos x * 0 – sin x * 1          [Since limh->0 (cos h – 1)/h = 0 and limh->0 (sin h)/h = 1]

= -sin x

So, f’(x) = -sin x

Question 11:

Find the derivative of the following functions:

(i) sin x cos x                   (ii) sec x                             (iii) 5 sec x + 4 cos x               (iv) cosec x

(v) 3cot x + 5cosec x     (vi) 5sin x – 6cos x + 7     (vii) 2tan x – 7sec x

(i) Let y = sin x cos x

Now, dy/dx = d(sin x cos x)/dx

=> dy/dx = d(sin x)/dx * cos x + sin x * d(cos x)/dx

=> dy/dx = cos x * cos x + sin x * (-sin x)                  [Since d(sin x)/dx = cos x, d(cos x)/dx = -sin x]

=> dy/dx = cos2 x – sin2 x

=> dy/dx = cos 2x                                                         [Since cos 2x = cos2 x – sin2 x]

(ii) Let f(x) = sec x

Now, from the first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{sec(x + h) – sec x}/h]

= limh->0 [{1/cos(x + h) – 1/cos x}/h]

= limh->0 {cos x - cos (x + h)}/[{cos x * cos (x + h) * h]

= (1/cos x) * limh->0 [{-2 * sin (x + x + h)/2 * sin (x - x - h)/2}/{h * cos (x + h)}]

= (1/cos x) * limh->0 [{-2 * sin (2x + h)/2 * sin (-h)/2}/{h * cos (x + h)}]

= (1/cos x) * limh->0 [{2 * sin (2x + h)/2 * sin (h)/2}/{h * cos (x + h)}]

= (1/cos x) * limh->0 [{sin (h)/2]/(h/2) * limh->0 [{sin (2x + h)/2}/cos (x + h)]

= (1/cos x) * 1 * (sin x/cos x)

= sec x * tan x

So, f’(x) = sec x * tan x

(iii) Let y = 5 sec x + 4 cos x

Now, dy/dx = d(5 sec x + 4 cos x)/dx

=> dy/dx = d(5 sec x)/dx + d(4 cos x)/dx

=> dy/dx = 5 * sec x * tan x – 4 * sin x

(iv) Let f(x) = cosec x

Now, from the first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{cosec(x + h) – cosec x}/h]

= limh->0 [{1/sin(x + h) – 1/sin x}/h]

= limh->0 {sin x - sin (x + h)}/[{sin x * sin (x + h) * h]

= (1/sin x) * limh->0 [{2 * cos(x + x + h)/2 * sin (x - x - h)/2}/{h * sin(x + h)}]

= (1/sin x) * limh->0 [{2 * cos(2x + h)/2 * sin (-h)/2}/{h * sin(x + h)}]

= -(1/sin x) * limh->0 [{2 * cos(2x + h)/2 * sin(h)/2}/{h * sin(x + h)}]

= -(1/sin x) * limh->0 [{sin (h)/2]/(h/2) * limh->0 [{cos(2x + h)/2}/sin(x + h)]

= -(1/sin x) * 1 * (cos x/sin x)

= -cosec x * cot x

So, f’(x) = -cosec x * cot x

(v) Let f(x) = 3cot x + 5cosec x

Now, df(x)/dx = d(3cot x + 5cosec x)/dx

=> f’(x) = d(3cot x)/dx + d(5cosec x)/dx

=> f’(x) = -3cosec2 x – 5cosec x * cot x                         [Since d(cot x)/dx = -cosec2 x]

(vi) Let f(x) = 5sin x – 6cos x + 7

Now, df(x)/dx = d(5sin x – 6cos x + 7)/dx

=> f’(x) = d(5sin x)/dx – d(6cos x)/dx + d(7)/dx

=> f’(x) = 5cos x – 6(-sin x) + 0

=> f’(x) = 5cos x + 6sin x

(vii) Let f(x) = 2tan x – 7sec x

Now, df(x)/dx = d(2tan x – 7sec x)/dx

=> f’(x) = d(2tan x)/dx – d(7sec x)/dx

=> f’(x) = 2sec2 x – 7sec x tan x               [Since d(tan x)/dx = sec2 x, d(sec x)/dx = sec x tan x]

Miscellaneous Exercise on Chapter 13

Question 1:

Find the derivative of the following functions from first principles:

(i) −x                            (ii) (-x)-1                     (iii) sin(x + 1)                          (iv) cos(x – π/8)

(i) Let f(x) = -x

Now, by first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{-(x + h) – (-x)}/h]

= limh->0 [{-x - h + x}/h]

= limh->0 [-h/h]

= limh->0 (-1)

= -1

(ii) Let f(x) = (-x)-1 = -1/(-x) = -1/x

Now, by first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{-1/(x + h) – (-1/x)}/h]

= limh->0 [{-1/(x + h) + 1/x}/{h * x (x + h)}]

= limh->0 [(-x + x + h)/{h * x (x + h)}]

= limh->0 [h/{h * x (x + h)}]

= limh->0 [1/{x(x + h)}]

= 1/(x * x)

= 1/x2

(iii) Let f(x) = sin(x + 1)

Now, by first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{sin(x + h + 1) – sin(x + 1)}/h]

= limh->0 [{2 * cos(x + h + 1 + x + 1)/2 * sin(x + h + 1 – x - 1)/2}/h]

= limh->0 [{2 * cos(2x + h + 2)/2 * sin(h/2)}/h]

= limh->0 [cos(2x + h + 2)/2 * sin(h/2)/(h/2)]

= limh->0 [cos(2x + h + 2)/2] * limh->0 [sin(h/2)/(h/2)]

= cos(2x + 0 + 2)/2 * 1                                                           [Since limx->0 (sin x)/x = 1]

= cos(x + 1)

(iv) Let f(x) = cos(x – π/8)

Now, by first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{cos(x + h - π/8) – cos(x - π/8)}/h]

= limh->0 [{-2 * sin(x + h - π/8 + x - π/8)/2 * sin(x + h - π/8 - x + π/8)/2}/h]

= limh->0 [{-2 * sin(2x + h - π/4)/2 * sin(h/2)}/h]

= -limh->0 [sin(2x + h - π/4)/2 * sin(h/2)/(h/2)]

= -limh->0 [sin(2x + h - π/4)/2] * limh->0 [sin(h/2)/(h/2)]

= -sin (2x + 0 - π/4)/2 * 1

= -sin (x - π/8)

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

Question 2:

(x + a)

Let f(x) = x + a

Now, by first principle,

f’(x) = limh->0 [{f(x + h) – f(x)}/h]

= limh->0 [{x + h + a – x - a}/h]

= limh->0 [h/h]

= limh->0 [1]

= 1

Question 3:

(px + q)(r/x + s)

Let f(x) = (px + q)(r/x + s)

Now, df(x)/dx = d[(px + q)(r/x + s)]/dx

=> f’(x) = (px + q) * d(r/x + s)/dx + (r/x + s) * d(px + q)/dx

= (px + q) * (-r/x2) + (r/x + s) * p

= -pr/x - qr/x2 + pr/x + ps

= ps – qr/x2

Question 4:

(ax + b)(cx + d)2

Let f(x) = (ax + b)(cx + d)2

Now, df(x)/dx = d[(ax + b)(cx + d)2]/dx

=> f’(x) = (ax + b) * d[(cx + d)2]/dx + (cx + d)2 * d(ax + b)/dx

= (ax + b) * 2 * (cx + d) * c + (cx + d)

= 2c(ax + b)(cx + d) + a(cx + d)2

Question 5:

(ax + b)/(cx + d)

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

Now, df(x)/dx = d[(ax + b)/(cx + d)]/dx

=> f’(x) = [(cx + d) * d(ax + b)/dx – (ax + b) * d(cx + d)/dx]/(cx + d)2

= [(cx + d) * a – (ax + b) * c]/(cx + d)2

= [acx + ad – acx - bc]/(cx + d)2

= (ad - bc)/(cx + d)2

Question 6:

(1 + 1/x)/(1 – 1/x)

Let f(x) = (1 + 1/x)/(1 – 1/x)

= {(x + 1)/x}/{(x – 1)/x}

= (x + 1)/(x - 1)

Now, df(x)/dx = d[(x + 1)/(x - 1)]/dx

=> f’(x) = {(x - 1) * d(x + 1)/dx - (x + 1) * d(x - 1)/dx}/(x - 1)2

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

= (x - 1 - x - 1)/(x - 1)2

= -2/(x - 1)2 , x ≠ 0, 1

Question 7:

1/(ax2 + bx + c)

Let f(x) = 1/(ax2 + bx + c)

Now, df(x)/dx = d[1/(ax2 + bx + c)]

=> f’(x) = {(ax2 + bx + c) * d(1)/dx – 1 * d(ax2 + bx + c)/dx}/(ax2 + bx + c)2

= {(ax2 + bx + c) * 0 – (2ax + b)}/(ax2 + bx + c)2

= -(2ax + b)/(ax2 + bx + c)2

Question 8:

(ax + b)/(px2 + qx + r)

Let f(x) = (ax + b)/(px2 + qx + r)

Now, df(x)/dx = d[(ax + b)/(px2 + qx + r)]/dx

=> f’(x) = {(px2 + qx + r) * d(ax + b)/dx – (ax + b) * d(px2 + qx + r)/dx}/(px2 + qx + r)2

= {(px2 + qx + r) * a – (ax + b) * (2px + q)}/(px2 + qx + r)2

= (apx2 + aqx + ar – 2apx2 – aqx – 2bpx - bq)}/(px2 + qx + r)2

= (-apx2 + ar – 2bpx - bq)}/(px2 + qx + r)2

Question 9:

(px2 + qx + r)/(ax + b)

Let f(x) = (px2 + qx + r)/(ax + b)

Now, df(x)/dx = d[(px2 + qx + r)/(ax + b)]/dx

=> f’(x) = {(ax + b) * d(px2 + qx + r)/dx - (px2 + qx + r) * d(ax + b)/dx}/(ax + b)2

= {(ax + b) * (2px + q) - (px2 + qx + r) * a}/(ax + b)2

= (2apx2 + aqx + 2bpx + bq - apx2 - aqx - ar)/(ax + b)2

= (apx2 + 2bpx + bq - ar)/(ax + b)2

Question 10:

a/x4 – b/x2 + cos x

Let f(x) = a/x4 – b/x2 + cos x

Now, df(x)/dx =  d[a/x4 – b/x2 + cos x]/dx

=> f’(x) = d[a/x4]/dx – d[b/x2]/dx + d[cos x]/dx

=> f’(x) = a * d[x-4]/dx – b * d[x-2]/dx + d[cos x]/dx

=> f’(x) = a * (-4 * x-5) – b * (-2 * x-3) + (-sin x)

=> f’(x) = -4a * x-5 – (-2b * x-3) - sin x

=> f’(x) = -4a/x5 + 2b/x3 - sin x

Question 11:

4√x - 2

Let f(x) = 4√x - 2

Now, df(x)/dx = d[4√x - 2]/dx

=> f’(x) = d[4√x]/dx – d[2]/dx

=> f’(x) = 4d[√x]/dx – 0

=> f’(x) = 4d[x1/2]/dx

=> f’(x) = 4 * [(1/2) * x1/2-1]

=> f’(x) = 2x-1/2

=> f’(x) = 2/x1/2

=> f’(x) = 2/√x

Question 12:

(ax + b)n

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

Now, df(x)/dx = d[(ax + b)n]

=> f’(x) = n * (ax + b)n-1 * d[(ax + b)]/dx            [Since d(xn)/dx = nxn-1]

=> f’(x) = n * (ax + b)n-1 * a

=> f’(x) = an(ax + b)n-1

Question 13:

(ax + b)n (cx + d)m

Let f(x) = (ax + b)n (cx + d)m

Now, df(x)/dx = d[(ax + b)n (cx + d)m]/dx

=> f’(x) = (ax + b)n * d[(cx + d)m ] + (cx + d)m * d[(ax + b)n ]

=> f’(x) = (ax + b)n * m * (cx + d)m-1 * d[(cx + d)]/dx + (cx + d)m * n * (ax + b)n-1 * d[(ax + b)]/dx

=> f’(x) = (ax + b)n * m * (cx + d)m-1 * c + (cx + d)n * n * (ax + b)n-1 * a

=> f’(x) = cm(ax + b)n * (cx + d)m-1 + an(cx + d)n * (ax + b)n-1

=> f’(x) = (ax + b)n-1 (cx + d)m-1 [cm(ax + b) + an(cx + d)]

Question 14:

sin(x + a)

Let f(x) = sin(x + a)

Now, df(x)/dx = d[sin(x + a)]/dx

=> f’(x) = cos(x + a) * d(x + a)/dx

=> f’(x) = cos(x + a) * 1

=> f’(x) = cos(x + a)

Question 15:

cosec x * cot x

Let f(x) = cosec x * cot x

Now, df(x)/dx = d(cosec x * cot x)/dx

=> f’(x) = d(cosec x)/dx * cot x + cosec x * d(cot x)/dx

=> f’(x) = (-cosec x * cot x) * cot x + cosec x * (-cosec2 x)

=> f’(x) = -cosec x * cot2 x – cosec3 x

Question 16:

cos x/(1 + sin x)

Let f(x) = cos x/(1 + sin x)

Now, df(x)/dx = d[cos x/(1 + sin x)]/dx

=> f’(x) = {(1 + sin x) * d(cos x)/dx – cos x * d(1 + sin x)/dx}/(1 + sin x)2

=> f’(x) = {(1 + sin x) * (-sin x) – cos x * (cos x)}/(1 + sin x)2

=> f’(x) = {-sin x – sin2 x – cos2 x}/(1 + sin x)2

=> f’(x) = {-sin x – (sin2 x + cos2 x)}/(1 + sin x)2

=> f’(x) = (-sin x – 1)/(1 + sin x)2           [Since sin2 x + cos2 x = 1]

=> f’(x) = -(1 + sin x)/(1 + sin x)2

=> f’(x) = -1/(1 + sin x)

Question 17:

(sin x + cos x)/(sin x - cos x)

Let f(x) = (sin x + cos x)/(sin x - cos x)

Now, df(x)/dx = d[(sin x + cos x)/(sin x - cos x)]/dx

f’(x) = {(sin x - cos x) * d[(sin x + cos x)]/dx – (sin x + cos x) * d[(sin x - cos x)]/dx}/(sin x - cos x)2

=> f’(x) = {(sin x - cos x) * (cos x - sin x) – (sin x + cos x) * (cos x + sin x)}/(sin x - cos x)2

=> f’(x) = {-(sin x - cos x) * (sin x - cos x) – (sin x + cos x)2}/(sin x - cos x)2

=> f’(x) = {-(sin x - cos x)2  – (sin x + cos x)2}/(sin x - cos x)2

=> f’(x) = -{sin2 x + cos2 x - 2 * sin x * cos x  + sin2 x + cos2 x + 2 * sin x * cos x}/(sin x - cos x)2

=> f’(x) = -{sin2 x + cos2 x - 2 * sin x * cos x  + sin2 x + cos2 x + 2 * sin x * cos x}/(sin x - cos x)2

=> f’(x) = -(1 + 1)/(sin x - cos x)2                 [Since sin2 x + cos2 x = 1]

=> f’(x) = -2/(sin x - cos x)2

Question 18:

(sec x - 1)/(sec x + 1)

Let f(x) = (sec x - 1)/(sec x + 1)

= (1/cos x - 1)/(1/cos x + 1)

= (1 – cos x)/(1 + cos x)

Now, df(x)/dx = d[(1 – cos x)/(1 + cos x)]/dx

=> f’(x) = {(1 + cos x) * d(1 - cos x)/dx - (1 - cos x) * d(1 + cos x)/dx}/(1 + cos x)2

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

= (sin x + sin x * cos x + sin x – sin x * cos x)/ (1 + cos x)2

= 2sin x/(1 + cos x)2

= 2sin x/(1 + 1/sec x)2

= 2sin x/{(sec x + 1)/sec x}2

= 2sin x/{(sec x + 1)2/sec2 x}

= (2sin x * sec2 x)/{(sec x + 1)2

= (2sin x * 1/cos2 x)/{(sec x + 1)2

= (2sin x/cos x * 1/cos x)/{(sec x + 1)2

= (2tan x * sec x)/{(sec x + 1)2

Question 19:

sinn x

Let f(x) = sinn x

Now,  df(x)/dx = d(sinn x)/dx

=> f’(x) = n * sinn-1 x * d(sin x)/dx                       [Since d(xn) = n * xn-1]

=> f’(x) = n * sinn-1 x * cos x

Question 20:

(a + b * sin x)/(c + d * cos x)

Let f(x) = (a + b * sin x)/(c + d * cos x)

Now, df(x)/dx = d[(a + b * sin x)/(c + d * cos x)]/dx

=> f’(x) = {(c + d * cos x) * d(a + b * sin x)/dx - (a + b * sin x) * d(c + d * cos x)/dx}/(c + d * cos x)2

= {(c + d * cos x) * b * cos x - (a + b * sin x) * (-d * sin x)}/(c + d * cos x)2

= (bc cos x + bd cos2 x + ad sin x + bd sin2 x)/(c + d * cos x)2

= {(bc cos x + ad sin x + bd (cos2 x + sin2 x)}/(c + d * cos x)2

= (bc cos x + ad sin x + bd)/(c + d * cos x)2                      [Since cos2 x + sin2 x = 1]

Question 21:

sin(x + a)/cos x

Let f(x) = sin(x + a)/cos x

Now, df(x)/dx = d[sin(x + a)/cos x]/dx

=> f’(x) = [cos x * d{sin(x + a)}/dx - sin(x + a) * d(cos x)/dx]/cos2 x

=> f’(x) =  [cos x * cos(x + a) - sin(x + a) * (-sin x)]/cos2 x

=> f’(x) =  [cos x * cos(x + a) + sin(x + a) * sin x]/cos2 x

=> f’(x) =  cos(x + a - x)/cos2 x                 [Apply cos(A + B) formula]

=> f’(x) =  cos a/cos2 x

Question 22:

x4(5 sin x – 3 cos x)

Let f(x) = x4(5 sin x – 3 cos x)

Now, df(x)/dx = d{x4(5 sin x – 3 cos x)}/dx

=> f’(x) = d(x4)/dx * (5 sin x – 3 cos x) + x4 * d(5 sin x – 3 cos x)/dx

=> f’(x) = 4x3(5 sin x – 3 cos x) + x4(5 cos x + 3 sin x)

=> f’(x) = x3(5x cos + 3x sin x + 20 sin x - 12 cos x)

Question 23:

(x2 + 1)cos x

Let f(x) = (x2 + 1) cos x

Now, df(x)/dx = d[(x2 + 1) cos x]/dx

=> f’(x) = (x2 + 1) * d(cos x)/dx + cos x * d(x2 + 1)/dx

=> f’(x) = (x2 + 1) * (-sin x) + cos x * 2x

=> f’(x) = -(x2 + 1) sin x + 2x cos x

Question 24:

(ax2 + sin x)(p + q cos x)

Let f(x) = (ax2 + sin x)(p + q cos x)

Now, df(x)/dx = d[(ax2 + sin x)(p + q cos x)]/dx

=> f’(x) = (ax2 + sin x) * d(p + q cos x)/dx + (p + q cos x) * d(ax2 + sin x)/dx

=> f’(x) = (ax2 + sin x) * (-q sin x) + (p + q cos x)(2ax + cos x)

=> f’(x) = -q sin x(ax2 + sin x) + (p + q cos x)(2ax + cos x)

Question 25:

(x + cos x)(x – tan x)

Let f(x) = (x + cos x)(x – tan x)

Now, df(x)/dx = d[(x + cos x)(x – tan x)]/dx

=> f’(x) = (x + cos x) * d(x – tan x)/dx + (x – tan x) * d(x + cos x)/dx

=> f’(x) = (x + cos x) * (1 – sec2 x) + (x – tan x) * (1 - sin x)               [Since d(tan x)/dx = sec2 x]

=> f’(x) = (x + cos x) * (-tan2 x) + (x – tan x) * (1 - sin x)             [Since 1 + tan2 x = sec2 x]

=> f’(x) = - tan2 x (x + cos x) + (x – tan x) * (1 - sin x)

Question 26:

(4x + 5 sin x)/(3x + 7 cos x)

Let f(x) = (4x + 5 sin x)/(3x + 7 cos x)

Now, df(x)/dx = d[(4x + 5 sin x)/(3x + 7 cos x)]/dx

=> f’(x) = [(3x + 7 cos x) * d(4x + 5 sin x)/dx - (4x + 5 sin x) * d(3x + 7 cos x)/dx]/(3x + 7 cos x)2

= [(3x + 7 cos x) * (4 + 5 cos x) - (4x + 5 sin x) * (3 - 7 sin x)]/(3x + 7 cos x)2

= [12x + 15x cos x + 28 cos x + 35 cos2 x – 12x + 28x sin x – 15 sin x + 35 sin2 x]/(3x + 7cos x)2

= [15x cos x + 28 cos x + 28x sin x – 15 sin x + 35 (sin2 x + cos2 x)]/(3x + 7cos x)2

= [15x cos x + 28 cos x + 28x sin x – 15 sin x + 35]/(3x + 7cos x)2

=> f’(x) = [35 + 15x cos x + 28 cos x + 28x sin x – 15 sin x]/(3x + 7cos x)2

Question 27:

{x2 cos(π/4)}/sin x

Let f(x) = {x2 cos(π/4)}/sin x

Now, df(x)/dx = d[{x2 cos(π/4)}/sin x]/dx

= cos(π/4) * d(x2/sin x)/dx

= cos(π/4) * [sin x * d(x2)/dx - x2 * d(sin x)/dx]/sin2 x

= cos(π/4) * [sin x * 2x - x2 * cos x]/sin2 x

= x cos(π/4) * [2 sin x – x cos x]/sin2 x

Question 28:

x/(1 + tan x)

Let f(x) = x/(1 + tan x)

Now, df(x)/dx = d[x/(1 + tan x)]/dx

= [(1 + tan x) * d(x)/dx - x * d(1 + tan x)/dx]/(1 + tan x)2

= [(1 + tan x) * 1 - x * sec2 x]/ (1 + tan x)2                 [Since d(tan x)/dx = sec2 x]

= [1 + tan x - x sec2 x]/ (1 + tan x)2

Question 29:

(x + sec x)(x – tan x)

Let f(x) = (x + sec x)(x – tan x)

Now, df(x)/dx = d{(x + sec x)(x – tan x)}/dx

=> f’(x) = (x + sec x) * d(x – tan x)/dx + (x – tan x) * d(x + sec x)/dx

= (x + sec x) * (1 – sec2 x) + (x – tan x) * (1 + sec x * tan x)

[Since d(tan x)/dx = sec2 x and d(sec x)/dx = sec x * tan x]

Question 30:

x/sinn x

Let f(x) = x/sinn x

Now, df(x)/dx = d{x/sinn x}/dx

=> f’(x) = {sinn x * d(x)/dx – x * d(sinn x)/dx}/(sinn x)2

=> f’(x) = {sinn x * 1 – x * n * sinn-1 x * d(sin x)/dx}/(sin2n x)

=> f’(x) = {sinn x – nx sinn-1 x * cos x}/(sin2n x)

=> f’(x) = sinn-1 x{sin x – nx * cos x}/(sin2n x)

=> f’(x) = (sin x – nx * cos x}/(sin2n x / sinn-1 x)

=> f’(x) = (sin x – nx * cos x}/(sin2n-n+1 x)

=> f’(x) = (sin x – nx * cos x}/(sinn+1 x)