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• NCERT Solutions for Class 11 Maths

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

• Last Updated : 30 Apr, 2021

### 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 sinn-1x cos x

For P(1),

P(1) = 1 sin1-1x cos x = cos x. Which is true.

n=k

n = k+1

Using the product rule, we have

(uv)’ = uv’+vu’

= (sink x)  + (sin x)

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

= (sink 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 nxn-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 nxn-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 nxn-1 and derivative of constant is 0.

f'(x) = (x2+1) (- sin x) + (cos x)[(2x2-1)+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 nxn-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 nxn-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 nxn-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 nxn-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 nxn-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 nxn-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 nxn-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

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