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

### Question 1. Find the derivative of x^{2} â€“ 2 at x = 10.

**Solution:**

f(x) = x

^{2}â€“ 2f(x+h) = (x+h)

^{2}â€“ 2From 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) x^{3} âˆ’ 27

**Solution:**

f(x) = x

^{3}â€“ 27f(x+h) = (x+h)

^{3}â€“ 27From the first principle,

f'(x) = 0

^{2}+3x(x+0)f'(x) = 3x

^{2}

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

**Solution:**

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

^{2}– 3x + 2f(x) = (x+h)

^{2}– 3(x+h) + 2From 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 x

^{n}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 x^{n} + ax^{n-1} + a^{2}x^{n-2} + ……………….+ a^{n-1}x + a^{n} for some fixed real number a.

**Solution:**

Given,

f(x) = x

^{n}+ ax^{n-1}+ a^{2}x^{n-2}+ ……………….+ a^{n-1}x + a^{n}As, the derivative of x

^{n}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) = x

^{2}– (a+b)x + abTaking derivative both sides,

As, the derivative of x

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

### (ii) (ax^{2} + b)^{2}

**Solution:**

f(x) = (ax

^{2}+ b)^{2}f(x) = (ax

^{2})^{2}+ 2(ax^{2})(b) + b^{2}Taking derivative both sides,

As, the derivative of x

^{n}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 x

^{n}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 x

^{n}is nx^{n-1}and derivative of constant is 0.f'(x) = (2x

^{0})-0f'(x) = 2

### (ii) (5x^{3} + 3x – 1)(x-1)

**Solution:**

f(x) = (5x

^{3}+ 3x – 1)(x-1)Taking derivative both sides,

Using product rule, we have

(uv)’ = uv’ + u’vAs, the derivative of x

^{n}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 x

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

### (iv) x^{5} (3-6x^{-9})

**Solution:**

f(x) = x

^{5}(3-6x^{-9})Taking derivative both sides,

Using product rule, we have

(uv)’ = uv’ + u’v

As, the derivative of x

^{n}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’vAs, the derivative of x

^{n}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 x

^{n}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 sinsinMultiplying 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 cossin

### (ii) sec x

**Solution:**

f(x) = sec x =

From the first principle,

Using the trigonometric identity,

cos a – cos b = -2 sinsinMultiply 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 cossinMultiply 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 (- cosec

^{2}x) + 5 (-cot x cosec x)f'(x) = – 3cosec

^{2}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 cossin

cos a – cos b = -2 sinsinMultiply 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) = sec

^{2}xSo, now

f'(x) = 2 g'(x) – 7

f'(x) = 2 (sec

^{2}x) – 7 (sec x tan x)f'(x) = 2sec

^{2}x – 7 sec x tan x