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• NCERT Solutions for Class 11 Maths

# Class 11 NCERT Solutions – Chapter 13 Limits And Derivatives – Exercise 13.2

• Last Updated : 30 Aug, 2022

### Question 1. Find the derivative of x2 – 2 at x = 10.

Solution:

f(x) = x2 – 2

f(x+h) = (x+h)2 – 2

From the first principle,

When, x = 10

f'(10) = 20 + 0

f'(10) = 20

### Question 2. Find the derivative of x at x = 1.

Solution:

f(x) = x

f(x+h) = x+h

From the first principle,

When, x = 1

f'(1) = 1

### Question 3. Find the derivative of 99x at x = l00.

Solution:

f(x) = 99x

f(x+h) = 99(x+h)

From the first principle,

When, x = 10

f'(100) = 99

### (i) x3 − 27

Solution:

f(x) = x3 – 27

f(x+h) = (x+h)3 – 27

From the first principle,

f'(x) = 02+3x(x+0)

f'(x) = 3x2

### (ii) (x-1) (x-2)

Solution:

f(x) = (x-1) (x-2) = x2 – 3x + 2

f(x) = (x+h)2 – 3(x+h) + 2

From the first principle,

f'(x) = 2x+0 – 3

f'(x) = 2x – 3

### (iii)

Solution:

From the first principle,

### (iv)

Solution:

From the first principle,

### Prove that f'(1) = 100 f'(0)

Solution:

Given,

By using this, taking derivative both sides

As, the derivative of xn is nxn-1 and derivative of constant is 0.

Now, then

Hence, we conclude that

f'(1) = 100 f'(0)

### Question 6. Find the derivative of xn + axn-1 + a2xn-2 + ……………….+ an-1x + an for some fixed real number a.

Solution:

Given,

f(x) = xn + axn-1 + a2xn-2 + ……………….+ an-1x + an

As, the derivative of xn is nxn-1 and derivative of constant is 0.

By using this, taking derivative both sides

### (i) (x-a) (x-b)

Solution:

f(x) = (x-a) (x-b)

f(x) = x2 – (a+b)x + ab

Taking derivative both sides,

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### (ii) (ax2 + b)2

Solution:

f(x) = (ax2 + b)2

f(x) = (ax2)2 + 2(ax2)(b) + b2

Taking derivative both sides,

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### (iii)

Solution:

Taking derivative both sides,

Using quotient rule, we have

### Question 8. Find the derivative of  for some constant a.

Solution:

Taking derivative both sides,

Using quotient rule, we have

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### (i)

Solution:

Taking derivative both sides,

As, the derivative of xn is nxn-1 and derivative of constant is 0.

f'(x) = (2x0)-0

f'(x) = 2

### (ii) (5x3 + 3x – 1)(x-1)

Solution:

f(x) = (5x3 + 3x – 1)(x-1)

Taking derivative both sides,

Using product rule, we have

(uv)’ = uv’ + u’v

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### (iii) x-3 (5+3x)

Solution:

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

Taking derivative both sides,

Using product rule, we have

(uv)’ = uv’ + u’v

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### (iv) x5 (3-6x-9)

Solution:

f(x) = x5 (3-6x-9)

Taking derivative both sides,

Using product rule, we have

(uv)’ = uv’ + u’v

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### (v) x-4 (3-4x-5)

Solution:

f(x) = x-4 (3-4x-5)

Taking derivative both sides,

Using product rule, we have

(uv)’ = uv’ + u’v

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### (vi)

Solution:

Taking derivative both sides,

Using quotient rule, we have

As, the derivative of xn is nxn-1 and derivative of constant is 0.

### Question 10. Find the derivative of cos x from first principle.

Solution:

Here, f(x) = cos x

f(x+h) = cos (x+h)

From the first principle,

Using the trigonometric identity,

cos a – cos b = -2 sin  sin

Multiplying and dividing by 2,

f'(x) = -sin (x) (1)

f'(x) = -sin x

### (i) sin x cos x

Solution:

f(x) = sin x cos x

f(x+h) = sin (x+h) cos (x+h)

From the first principle,

Using the trigonometric identity,

sin A cos B = (sin (A+B) + sin(A-B))

Using the trigonometric identity,

sin A – sin B = 2 cos  sin

### (ii) sec x

Solution:

f(x) = sec x =

From the first principle,

Using the trigonometric identity,

cos a – cos b = -2 sin  sin

Multiply and divide by 2, we have

### (iii) 5 sec x + 4 cos x

Solution:

f(x) = 5 sec x + 4 cos x

Taking derivative both sides,

f'(x) = 5 (tan x sec x) + 4 (-sin x)

f'(x) = 5 tan x sec x – 4 sin x

### (iv) cosec x

Solution:

f(x) = cosec x =

From the first principle,

Using the trigonometric identity,

sin a – sin b = 2 cos  sin

Multiply and divide by 2, we have

### (v) 3 cot x + 5 cosec x

Solution:

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

Taking derivative both sides,

f'(x) = 3 g'(x) + 5

Here,

g(x) = cot x =

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

So, now

f'(x) = 3 g'(x) + 5

f'(x) = 3 (- cosec2 x) + 5 (-cot x cosec x)

f'(x) = – 3cosec2 x – 5 cot x cosec x

### (vi) 5 sin x – 6 cos x + 7

Solution:

f(x) = 5 sin x – 6 cos x + 7

f(x+h) = 5 sin (x+h) – 6 cos (x+h) + 7

From the first principle,

Using the trigonometric identity,

sin a – sin b = 2 cos  sin

cos a – cos b = -2 sin  sin

Multiply and divide by 2, we get

f'(x) = 5 cos x (1) + 6  sin x (1)

f'(x) = 5 cos x + 6  sin x

### (vii) 2 tan x – 7 sec x

Solution:

f(x) = 2 tan x – 7 sec x

Taking derivative both sides,

f'(x) =

f'(x) = 2 g'(x) – 7

Here,

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

So, now

f'(x) = 2 g'(x) – 7

f'(x) = 2 (sec2x) – 7 (sec x tan x)

f'(x) = 2sec2x – 7 sec x tan x

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