RD Sharma Solutions for Class 12 Maths Chapter 11 Differentiation

RD Sharma books offer several questions for practice at the end of each chapter. RD Sharma solutions provided here are easily readable and sketched in such a way to help students clear all their doubts that they might face, while answering the given problems in exercises. These solutions are prepared by BYJU’S experts in Maths. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 11 Differentiation is provided here. Students can refer and download Chapter 11 Differentiation from the given links. These solutions facilitate students to create good knowledge about basic concepts of Mathematics. This chapter is based on the differentiation of given function. Some of the essential topics of this chapter are listed below.

• Recapitulation of the product rule, quotient rule and differentiation of a constant with an illustration.
• Differentiation of inverse trigonometric functions from first principles.
• Differentiation of a function of a function.
• Differentiation of inverse trigonometric functions by the chain rule.
• Differentiation by using trigonometrical substitutions.
• Differentiation of implicit functions.
• Logarithmic differentiation.
• Differentiation of infinite series.
• Differentiation of parametric functions.
• Differentiation of a function with respect to another function.
• Differentiation of determinants.

RD Sharma Solutions For Class 12 Maths Chapter 11 Differentiation:-

Access answers to Maths RD Sharma Solutions For Class 12 Chapter 11 – Differentiation

Exercise 11.1 Page No: 11.17

Differentiate the following functions from the first principles:

1. e-x

Solution:

2. e3x

Solution:

3. eax + b

Solution:

4. ecos x

Solution:

We have to find the derivative of ecos x with the first principle method,

So, let f (x) = ecos x

By using the first principle formula, we get,

Solution:

Exercise 11.2 Page No: 11.37

Differentiate the following functions with respect to x:

1. Sin (3x + 5)

Solution:

Given Sin (3x + 5)

2. tan2 x

Solution:

Given tan2 x

3. tan (xo + 45o)

Solution:

Let y = tan (x° + 45°)

First, we will convert the angle from degrees to radians.

4. Sin (log x)

Solution:

Given sin (log x)

Solution:

6. etan x

Solution:

7. Sin2 (2x + 1)

Solution:

Let y = sin2 (2x + 1)

On differentiating y with respect to x, we get

8. log7 (2x – 3)

Solution:

9. tan 5xo

Solution:

Let y = tan (5x°)

First, we will convert the angle from degrees to radians. We have

Solution:

Solution:

12. logx 3

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

18. (log sin x)2

Solution:

Let y = (log sin x)2

Solution:

Solution:

21. e3x cos 2x

Solution:

22. Sin (log sin x)

Solution:

23. etan 3x

Solution:

Solution:

Solution:

Solution:

27. tan (esin x)

Solution:

Solution:

Solution:

30. log (cosec x – cot x)

Solution:

Solution:

Solution:

33. tan-1 (ex)

Solution:

Solution:

35. sin (2 sin-1 x)

Solution:

Let y = sin (2sin–1x)

On differentiating y with respect to x, we get

Solution:

Solution:

Exercise 11.3 Page No: 11.62

Differentiate the following functions with respect to x:

Solution:

Solution:

Solution:

Let,

Solution:

Let,

Solution:

Solution:

7. Sin-1 (2x2 – 1), 0 < x < 1

Solution:

8. Sin-1 (1 – 2x2), 0 < x < 1

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Let,

Solution:

Solution:

Solution:

Solution:

Exercise 11.4 Page No: 11.74

Find dy/dx in each of the following:

1. xy = c2

Solution:

2. y3 – 3xy2 = x3 + 3x2y

Solution:

Given y3 – 3xy2 = x3 + 3x2y,

Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,

3. x2/3 + y2/3 = a2/3

Solution:

Given x2/3 + y2/3 = a2/3,

Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,

4. 4x + 3y = log (4x – 3y)

Solution:

Given 4x + 3y = log (4x – 3y),

Now we have to find dy/dx of it, so by differentiating the equation on both sides with respect to x, we get,

Solution:

6. x5 + y5 = 5xy

Solution:

Given x5 + y5 = 5xy

Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,

7. (x + y)2 = 2axy

Solution:

Given (x + y)2 = 2axy

Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,

8. (x2 + y2)2 = xy

Solution:

Given (x + y)2 = 2axy

Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,

9. Tan-1 (x2 + y2)

Solution:

Given tan – 1(x2 + y2) = a,

Now we have to find dy/dx of given function, so by differentiating the equation on both sides with respect to x, we get,

Solution:

11. Sin xy + cos (x + y) = 1

Solution:

Given Sin x y + cos (x + y) = 1

Now we have to find dy/dx of given function, so by differentiating the equation on both sides with respect to x, we get,

Solution:

Solution:

Solution:

Solution:

Exercise 11.5 Page No: 11.88

Differentiate the following functions with respect to x:

1. x1/x

Solution:

2. xsin x

Solution:

3. (1 + cos x)x

Solution:

Solution:

5. (log x)x

Solution:

6. (log x)cos x

Solution:

Let y = (log x)cos x

Taking log both the sides, we get

7. (Sin x)cos x

Solution:

8. ex log x

Solution:

9. (Sin x)log x

Solution:

10. 10log sin x

Solution:

11. (log x)log x

Solution:

Solution:

13. Sin (xx)

Solution:

14. (Sin-1 x)x

Solution:

Solution:

16. (tan x)1/x

Solution:

Solution:

18. (i) (xx) √x

Solution:

Solution:

Solution:

Solution:

Solution:

18. (vi) esin x + (tan x)x

Solution:

18. (vii) (cos x)x + (sin x)1/x

Solution:

Solution:

19. y = ex + 10x + xx

Solution:

20. y = xn + nx + xx + nn

Solution:

Exercise 11.6 Page No: 11.98

Solution:

Solution:

Solution:

Solution:

Exercise 11.7 Page No: 11.103

Find dy/dx, when

1. x = at2 and y = 2 at

Solution:

2. x = a (θ + sin θ) and y = a (1 – cos θ)

Solution:

3. x = a cos θ and y = b sin θ

Solution:

Given x = a cos θ and y = b sin θ

4. x = a eθ (sin θ – cos θ), y = a eθ (sin θ + cos θ)

Solution:

5. x = b sin2 θ and y = a cos2 θ

Solution:

6. x = a (1 – cos θ) and y = a (θ + sin θ) at θ = π/2

Solution:

Solution:

Solution:

9. x = a (cos θ + θ sin θ) and y = a (sin θ – θ cos θ)

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Exercise 11.8 Page No: 11.112

1. Differentiate x2 with respect to x3.

Solution:

2. Differentiate log (1 +x2) with respect to tan-1 x.

Solution:

3. Differentiate (log x)x with respect to log x.

Solution:

4. Differentiate sin-1 √ (1-x2) with respect to cos-1x, if

(i) x ∈ (0, 1)

(ii) x ∈ (-1, 0)

Solution:

(i) Given sin-1 √ (1-x2)

(ii) Given sin-1 √ (1-x2)

Solution:

(i) Let

(ii) Let

(iii) Let

Solution:

(i) x ∈ (0, 1/ √2)

(ii) x ∈ (1/√2, 1)

Solution:

(i) Let

(ii) Let

8. Differentiate (cos x)sin x with respect to (sin x)cos x.

Solution:

Solution:

Solution:

Also, access RD Sharma Solutions for Class 12 Maths Chapter 11 Differentiation

Exercise 11.1 Solutions

Exercise 11.2 Solutions

Exercise 11.3 Solutions

Exercise 11.4 Solutions

Exercise 11.5 Solutions

Exercise 11.6 Solutions

Exercise 11.7 Solutions

Exercise 11.8 Solutions