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

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

Question 4. Find the derivative of the following functions from first principle.

(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,

Question 5. For the function

f(x) = 

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

Solution:

Given,

By using this, taking derivative both sides

As, the derivative of xⁿ is nx^n-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 + ax^n-1 + a²x^n-2 + ……………….+ a^n-1x + aⁿ for some fixed real number a.

Solution:

Given,

f(x) = xⁿ + ax^n-1 + a²x^n-2 + ……………….+ a^n-1x + aⁿ

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

By using this, taking derivative both sides

Question 7. For some constants a and b, find the derivative of

(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 nx^n-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 nx^n-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 nx^n-1 and derivative of constant is 0.

Question 9. Find the derivative of

(i) 

Solution:

Taking derivative both sides,

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

(vi) 

Solution:

Taking derivative both sides,

Using quotient rule, we have

As, the derivative of xⁿ is nx^n-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

Question 11. Find the derivative of the following functions:

(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