Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Miscellaneous Exercise on Chapter 13 | Set 2

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 16: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

Question 17: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

Question 18: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

Question 19: sinⁿ x

Solution:

f(x) = sinⁿ x

When n = 1,

f(x) = sin x

When n = 2,

f(x) = sin² x = sin x sin x 

Using the product rule, we have

(uv)’ = uv’+vu’

f'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

f(x) = sin³ x = sin² x sin x

Using the product rule, we have

(uv)’ = uv’+vu’

Pattern w.r.t n is seen here, as follows

Let’s check this statement.

For P(n) = n sin^n-1x cos x

For P(1),

P(1) = 1 sin^1-1x cos x = cos x. Which is true.

n=k

n = k+1

Using the product rule, we have

(uv)’ = uv’+vu’

= (sin^k x)  + (sin x) 

= (sin^k x) (cos x) + (sin x) (k sin^k-1 x cos x)

= (sin^k x) (cos x)[k+1]

Hence proved for P(k+1).

So, is true.

Question 20: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Question 21: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

Let’s take g(x) = sin (x+a)

g(x+h) = sin((x+h)+a)

From the first principle,

Using the trigonometric identity,

sin A – sin B = 2 cos   sin 

Multiply and divide by 2, we have

Hence, 

Using the trigonometric identity,

cos A cos B + sin A sin B = cos (A-B)

Question 22: x⁴(5sin x – 3cos x)

Solution:

f(x) = x⁴ (5sin x – 3cos x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

f'(x) = (x⁴) [(5 cos x) – (3 (- sin x))] + (5sin x – 3cos x)(4x³)

f'(x) = (x⁴) [(5 cos x) + (3 sin x)] + (5sin x – 3cos x)(4x³)

f'(x) = (x³) [5x cos x + 3x sin x + 20sin x – 12 cos x]

Question 23: (x²+1) cos x

Solution:

f(x) = (x²+1) cos x

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

f'(x) = (x²+1) (- sin x) + (cos x)[(2x^2-1)+0]

f'(x) = -x² sin x- sin x + 2x cos x

Question 24: 

Solution:

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Question 25: (x + cos x)(x – tan x)

Solution:

f(x) = (x + cos x)(x – tan x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

g'(x) = sec²x

Hence, 

f'(x) = (x + cos x)  + (x – tan x)[1 + (- sin x)]

f'(x) = (x + cos x) [1 – (sec² x)] + (x – tan x)[1 – sin x]

f'(x) = (x + cos x) [tan² x] + (x – tan x)[1 – sin x]

f'(x) = tan² x(x + cos x) + (x – tan x)[1 – sin x]

Question 26: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Question 27: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Question 28: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

g'(x) = sec²x

Hence, 

Question 29: (x + sec x) (x-tan x)

Solution:

f(x) = (x + sec x) (x-tan x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

g'(x) = sec²x

Now, let’s take h(x) = sec x = 

h(x+h) =

From the first principle,

Using the trigonometric identity,

cos a – cos b = -2 sin   sin 

Multiply and divide by 2, we have

Hence, 

Question 30: 

Solution:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take, g(x) = sinn x

When n = 1,

g(x) = sin x

When n = 2,

Using the product rule, we have

(uv)’ = uv’+vu’

g'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

g(x) = sin³ x = sin² x sin x

Using the product rule, we have

(uv)’ = uv’+vu’

Pattern w.r.t n is seen here, as follows

Let’s check this statement.

For 

For P(1),

. Which is true.

n=k

n = k+1

Using the product rule, we have

(uv)’ = uv’+vu’

Hence proved for P(k+1).

So,  is true.

So, the given equation will be