RBSE Class 12 Maths Chapter 7 – Differentiation Important questions and solutions are provided here.

## Table of Contents

Chapter 7 of RBSE Class 12 has six exercises, all of them deal with differentiating the given functions of various types. In this chapter, students will learn differentiating various functions such as algebraic expressions, trigonometric functions, logarithm functions, exponential function and so on. All these exercises will help students to work on various word problems.

- RBSE solution for Class 12 maths Chapter 8 | Application of Derivatives Important Questions and Solutions
- RBSE solution for Class 12 maths Chapter 7 | Differentiation Important Questions and Solutions
- Rbse solution for Class 12 maths Chapter 6 | Continuity and Differentiability Important Questions and Solutions
- RBSE solution for Class 12 maths Chapter 5 | Inverse of a Matrix and Linear Equations solutions
- Rbse solutions for class 12 maths chapter 4 Determinants
- Rbse solution for Class 12 Maths Chapter 3: Matrix
- Rbse solution for Class 12 Maths Chapter 2 Inverse Circular Functions
- Rbse solution for class 12 maths chapter 1 composite functions

### RBSE Maths Chapter 7: Exercise 7.1 Textbook Important Questions and Solutions

**Question 1: Differentiate the following function with respect to x.**

**tan(2x + 3)**

**Solution:**

Let the given function be y = tan (2x + 3)

Differentiating with respect to x,

dy/dx = d/dx (tan(2x + 3))

= sec^{2}(2x + 3) d/dx(2x + 3)

= sec^{2}(2x + 3).(2 × 1 + 0)

= 2 sec^{2}(2x + 3)

**Question 2: Differentiate the following function with respect to x.**

**sin{cos(x ^{2})}**

**Solution:**

Let the given function be y = sin {cos (x^{2})}

Differentiating with respect to x,

dy/dx = d/dx sin [cos(x^{2})]

= cos(cos x^{2}). (- sin x^{2}) d/dx (x^{2})

= -cos(cos x^{2}) (sin x^{2}). 2x

= -2x (sin x^{2}) cos(cos x^{2})

### RBSE Maths Chapter 7: Exercise 7.2 Textbook Important Questions and Solutions

**Question 3: Differentiate the following function with respect to x.**

**(a) sin ^{-1}{2x √(1 – x^{2})}, -1/√2 < x < 1/√2**

**(b) sin ^{-1}(3x – 4x^{3}) x ∈ (-1/2, 1/2)**

**Solution:**

(a) sin^{-1}{2x √(1 – x^{2})}, -1/√2 < x < 1/√2

Let y = sin^{-1}{2x √(1 – x^{2})}

Substitute x = sin θ

y = sin^{-1}{2 sin θ √(1 – sin^{2}θ)}

= sin^{-1}(2 sin θ cos θ)

= sin^{-1}(sin 2θ)

= 2θ

y = 2 sin^{-1}(x)

Differentiate with respect to x,

dy/dx = d/dx (2 sin^{-1}x)

2 d/dx(sin^{-1}x)

= 2[1/√(1 – x^{2})]

dy/dx = 2/√(1 – x^{2})

(b) sin^{-1}(3x – 4x^{3}) x ∈ (-1/2, 1/2)

Let y = sin^{-1}(3x – 4x^{3})

Substituting x = sin θ,

θ = sin^{-1}x

Thus, y = sin^{-1}(3 sin θ – 4 sin^{3}θ)

= sin^{-1}(sin 3θ)

= 3θ

= 3 sin^{-1}x

Differentiate with respect to x,

dy/dx = d/dx (3 sin^{-1}x)

= 3 d/dx (sin^{-1}x)

= 3[1/√(1 – x^{2})]

dy/dx = 3/√(1 – x^{2})

**Question 4: Differentiate the following function with respect to x.**

**(a) sec-1[1/(2x ^{2} – 1)]; x ∈ (0, 1/√2)**

**(b) cos-1[(1 – x ^{2})/ (1 + x^{2})]; x ∈ (0, ∞)**

**Solution:**

(a) sec-1[1/(2x^{2} – 1)]; x ∈ (0, 1/√2)

Let y = sec^{-1}[1/(2x^{2} – 1)]

Substituting x = cos θ,

y = sec^{-1}[1/(2 cos^{2}θ – 1)]

= sec^{-1}(1/ cos 2θ)

= sec^{-1}(sec 2θ)

y = 2θ

y = 2 cos^{-1}x

Differentiating with respect to x,

dy/dx = d/dx (2 cos^{-1}x)

= 2 d/dx(cos^{-1}x)

= 2[-1/√(1 – x^{2})]

dy/dx = -2/√(1 – x^{2})

(b) cos-1[(1 – x^{2})/ (1 + x^{2})]; x ∈ (0, ∞)

Let y = cos^{-1}[(1 – x^{2})/ (1 + x^{2})]

Substituting x = tan θ

y = cos^{-1}[(1 – tan^{2}θ)/ (1 + tan^{2}θ)

= cos^{-1}(cos 2θ)

y = 2θ

y = 2 tan^{-1}x

Differentiating with respect to x,

dy/dx = d/dx(2 tan^{-1}x)

= 2[1/(1 + x^{2})]

dy/dx = 2/(1 + x^{2})

### RBSE Maths Chapter 7: Exercise 7.3 Textbook Important Questions and Solutions

**Question 5: Find dy/dx.**

**(a) 2x + 3y = sin y**

**(b) x ^{2} + xy + y^{2} = 200**

**Solution:**

(a) 2x + 3y = sin y

Differentiate with respect to x on both sides,

d/dx(2x) + d/dx(3y) = d/dx(sin y)

2(1) + 3(dy/dx) = cos y (dy/dx)

2 = (cosy – 3) dy/dx

Therefore, dy/dx = 2/(cos y – 3)

(b) x^{2} + xy + y^{2} = 200

Differentiate with respect to x on both sides,

d/dx (x^{2}) + d/dx (xy) + d/dx (y^{2}) = d/dx(200)

2x + x(dy/dx) + 2y (dy/dx) = 0

(x + 2y)dy/dx = -2x – y

Therefore, dy/dx = -(2x + y)/(x + 2y)

**Question 6: Find dy/dx.**

**(a) √x + √y = √a**

**(b) tan(x + y) + tan(x – y) = 4**

**Solution:**

(a) √x + √y = √a

Differentiating with respect to x,

d/dx(√x) + d/dx(√y) = d/dx(√a)[1/(2√x)].1 + [1/(2√y)] dy/dx = 0[1/(2√y)] dy/dx = -1/(2√x)

dy/dx = -(√y)/(√x)

Therefore, dy/dx = -√(y/x)

(ii) tan(x + y) + tan(x – y) = 4

Differentiating with respect to x,

d/dx [tan(x + y)] + d/dx [tan(x – y)] = d/dx (4)

sec^{2}(x + y) d/dx(x + y) + sec^{2}(x – y)d/dx(x – y) = 0

sec^{2}(x + y) [1 + (dy/dx)] + sec^{2}(x – y)[1 – (dy/dx)] = 0

sec^{2}(x + y) + sec^{2}(x + y)dy/dx + sec^{2}(x – y) – sec^{2}(x – y)dy/dx = 0[sec^{2}(x + y) – sec^{2}(x – y)]dy/dx = -sec^{2}(x + y) – sec^{2}(x – y)

-[sec^{2}(x + y) – sec^{2}(x + y)] dy/dx = -[sec^{2}(x + y) + sec^{2}(x – y)]

Therefore, dy/dx = [sec^{2}(x + y) + sec^{2}(x – y)]/ [sec^{2}(x – y) – sec^{2}(x + y)]

### RBSE Maths Chapter 7: Exercise 7.4 Textbook Important Questions and Solutions

**Question 7: Find dy/dx, when x = log t, y = e ^{t} + cos t.**

**Solution:**

Given,

x = log t, y = e^{t} + cos t

Consider, x = log t

Differentiating with respect to t on both sides,

dx/dt = d/dt (log t)

dx/dt = 1/t

Now, y = e^{t} + cos t

Differentiating with respect to t on both sides,

dy/dt = d/dt(e^{t}) + d/dt(cos t)

dy/dt = e^{t} – sin t

dy/dt = (dy/dt)/ (dx/dt)

= (e^{t} – sin t)/ (1/t)

= t(e^{t} – sin t)

Therefore, dy/dt = t(e^{t} – sin t)

**Question 8: Find dy/dx, when x = a cos θ, y = b sin θ.**

**Solution:**

Given,

x = a cos θ, y = b sin θ

Consider, x = a cos θ

Differentiating with respect to θ on both sides,

dx/dθ = d/dθ(a cos θ)

dx/dθ = -a sin θ

Now, y = b sin θ

Differentiating with respect to θ on both sides,

dy/dθ = d/dθ(b sin θ)

dy/dθ = b cos θ

dy/dx = (dy/dθ)/ (dx/dθ)

= (b cos θ)/ (-a sin θ)

= -(b/a) cot θ

Therefore, dy/dx = -(b/a) cot θ

### RBSE Maths Chapter 7: Exercise 7.5 Textbook Important Questions and Solutions

**Question 9: Find d ^{2}y/dx^{2}, when y = x^{2} + 3x + 2.**

**Solution:**

Given,

y = x^{2} + 3x + 2

Differentiating with respect to x on both sides,

dy/dx = d/dx(x^{2} + 3x + 2)

dy/dx = 2x + 3

Again, differentiating with respect to x on both sides,

d/dx(dy/dx) = d/dx(2x + 3)

d^{2}y/dx^{2} = 2(1) + 0

Therefore, d^{2}y/dx^{2} = 2

**Question 10: If y = a cos nx + b sin nx, then prove that d ^{2}y/dx^{2} + n^{2}y = 0.**

**Solution:**

Given,

y = a cos nx + b sin nx

Differentiating with respect to x on both sides,

dy/dx = d/dx(a cos nx + b sin nx)

= -na sin nx + nb cos nx

Again, differentiating with respect to x on both sides,

d/dx(dy/dx) = d/dx(-na sin nx + nb cos nx)

d^{2}y/dx^{2} = -n^{2}a cos nx – n^{2}b sin nx

= -n^{2}(a cos nx + b sin nx)

= -n^{2}y (from the given)

Therefore, d^{2}y/dx^{2} + n^{2}y = 0

Hence proved.

### RBSE Maths Chapter 7: Exercise 7.6 Textbook Important Questions and Solutions

**Question 11: Verify Rolle’s theorem for the functions given below.**

**f(x) = cos 2x; x ∈ [0, π]**

**Solution:**

Given function

f(x) = cos 2x, x ∈ [0, π]

⇒ f(x) = cos 2x is defined in the interval [0, π].

We know that the cosine is continuous in this domain.

Thus, it is continuous in [0, π].

Therefore, f'(x) = -2 sin 2x exists wherein x ∈ (0, π)

⇒ f(x) is differentiable in (0, π).

Now,

f(0) = cos 0 = 1

f(π) = cos 2π = 1

∴ f(0) = f(π) = 1

Thus, all the conditions of Rolle’s theorem are satisfied hence, at least one point c such that c ∈ (0, π) and f'(c) = 0.

f'(c) = 0

-2 sin 2c = 0

sin 2c = 0

2c = π

c = π/2 is an element of (0,π)

∴ c = π/2 ∈ (0,π) such that f'(c) = 0

Hence, for c = π/2, Rolle’s theorem is satisfied.

**Question 12: Verify Lagrange’s mean value theorem for the function given below.**

**f(x) = x ^{2} – 3x + 2; x ∈ [-2, 3]**

**Solution:**

Given function is:

f(x) = x2 – 3x + 2, x ∈ [- 2, 3]

Thus, f(x) is continuous in the interval [-2, 3].

Hence, f'(x) is finite and exists in (-2, 3).

That means, f(x) is differentiable in (- 2, 3).

Therefore, f(x) satisfies both the conditions of Langrange’s mean value theorem.

f'(c) = [f(b) – f(a)]/(b – a)

2c – 3 = [f(3) – f(-2)]/ [3 – (-2)]

2c – 3 = (2 – 12)/5

2c – 3 = -10/2

2c – 3 = -2

2c = -2 + 3

2c = 1

c = 1/2

Thus, c = 1/2 ∈ (-2, 3)

Hence, the Langrange’s mean value theorem is satisfied.

### RBSE Maths Chapter 7: Additional Important Questions and Solutions

**Question 1: Differentiate log (x/a ^{x}) with respect to x.**

**Solution:**

Let the given function be:

y = log(x/ax)

y = log x – log ax

y = log x – x log a

Differentiating with respect to x on both sides,

dy/dx = d/dx(log x) – d/dx(x log a)

dy/dx = (1/x) – log a (1)

Therefore, dy/dx = (1/x) – log a

**Question 2: If sin y = x sin(a + y), then prove that dy/dx = sin ^{2}(a + y)/sin a.**

**Solution:**

Given function is:

sin y = x sin(a + y)

x = sin y/sin(a + y)

Differentiating with respect to y on both sides,

dx/dy = [sin(a + y). cos y – sin y. cos(a + y)]/ sin^{2}(a + y)

dx/dy = [sin(a + y – y)]/ sin^{2}(a + y)

dx/dy = sin a/sin^{2}(a + y)

Therefore, dy/dx = sin^{2}(a + y)/sin a