# RD Sharma Solutions for Class 11 Maths Chapter 30 Derivatives

## RD Sharma Solutions Class 11 Maths Chapter 30 â€“ Download Free PDF Updated Session for 2021-22

RD Sharma Solutions for Class 11 Maths Chapter 30 – DerivativesÂ are provided here for students to score good marks in the board exams. RD Sharma Class 11 Chapter 30 has explanatory answers to exercise wise problems in a comprehensive manner. Chapter 30 provides students with the knowledge of derivatives at a point.

The PDF of RD Sharma Solutions is made available both chapter wise and exercise wise to help students ace the examination. Students who aim to perform well in the board exams are advised to solve problems using the solutions PDF. The main aim of creating solutions is to help students clear their doubts and improve conceptual knowledge. RD Sharma Class 11 Maths Solutions, free PDF links are available below.

Chapter 30 – Derivatives contains five exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

• Derivative at a point.
• Physical interpretation of derivative at a point.
• Geometrical interpretation of derivative at a point.
• Derivative of a function.
• Derivative as a rate measurer.
• Differentiation from first principles.
• Fundamental rules for differentiation.
• Product rule for differentiation.
• Quotient rule for differentiation.

## Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 30 – Derivatives

### Access answers to RD Sharma Solutions for Class 11 Maths Chapter 30 – Derivatives

#### EXERCISE 30.1 PAGE NO: 30.3

1. Find the derivative of f(x) = 3x at x = 2

Solution:

Given:

f(x) = 3x

By using the derivative formula,

2. Find the derivative of f(x) = x2Â â€“ 2 at x = 10

Solution:

Given:

f(x) = x2Â â€“ 2

By using the derivative formula,

= 0 + 20 = 20

Hence,

Derivative of f(x) = x2Â â€“ 2 at x = 10 is 20

3. Find the derivative of f(x) = 99x at x = 100.

Solution:

Given:

f(x) = 99x

By using the derivative formula,

4. Find the derivative of f(x) = x at x = 1

Solution:

Given:

f(x) = x

By using the derivative formula,

5. Find the derivative of f(x) = cos x at x = 0

Solution:

Given:

f(x) = cos x

By using the derivative formula,

6. Find the derivative of f(x) = tan x at x = 0

Solution:

Given:

f(x) = tan x

By using the derivative formula,

7. Find the derivatives of the following functions at the indicated points:
(i) sin x at x = Ï€/2
(ii) x at x = 1

(iii) 2 cos x at x = Ï€/2

(iv) sin 2xat x = Ï€/2

Solution:

(i) sin x at x = Ï€/2

Given:

f (x) = sin x

By using the derivative formula,

[Since it is of indeterminate form. Let us try to evaluate the limit.]

We know that 1 â€“ cos x = 2 sin2(x/2)

(ii) x at x = 1

Given:

f (x) = x

By using the derivative formula,

(iii) 2 cos x at x = Ï€/2

Given:

f (x) = 2 cos x

By using the derivative formula,

(iv) sin 2xat x = Ï€/2

Solution:

Given:

f (x) = sin 2x

By using the derivative formula,

[Since it is of indeterminate form. We shall apply sandwich theorem to evaluate the limit.]

Now, multiply numerator and denominator by 2, we get

#### EXERCISE 30.2 PAGE NO: 30.25

1. Differentiate each of the following from first principles:
(i) 2/x
(ii) 1/âˆšx

(iii) 1/x3

(iv) [x2 + 1]/ x

(v) [x2 – 1] / x

Solution:

(i) 2/x

Given:

f (x) = 2/x

By using the formula,

âˆ´ Derivative of f(x) = 2/x is -2x-2

(ii) 1/âˆšx

Given:

f (x) = 1/âˆšx

By using the formula,

âˆ´ Derivative of f(x) = 1/âˆšx is -1/2 x-3/2

(iii) 1/x3

Given:

f (x) = 1/x3

By using the formula,

âˆ´ Derivative of f(x) = 1/x3 is -3x-4

(iv) [x2 + 1]/ x

Given:

f (x) = [x2 + 1]/ x

By using the formula,

= 1 â€“ 1/x2

âˆ´ Derivative of f(x) =Â 1 â€“ 1/x2

(v) [x2 – 1] / x

Given:

f (x) = [x2 – 1]/ x

By using the formula,

2. Differentiate each of the following from first principles:

(i) e-x

(ii) e3x

(iii) eax+b

Solution:

(i) e-x

Given:

f (x) = e-x

By using the formula,

(ii) e3x

Given:

f (x) = e3x

By using the formula,

(iii) eax+b

Given:

f (x) = eax+b

By using the formula,

3. Differentiate each of the following from first principles:

(i) âˆš(sin 2x)

(ii) sin x/x

Solution:

(i) âˆš(sin 2x)

Given:

f (x) = âˆš(sin 2x)

By using the formula,

(ii) sin x/x

Given:

f (x) = sin x/x

By using the formula,

4. Differentiate the following from first principles:

(i) tan2 x

(ii) tan (2x + 1)

Solution:

(i) tan2 x

Given:

f (x) = tan2 x

By using the formula,

(ii) tan (2x + 1)

Given:

f (x) = tan (2x + 1)

By using the formula,

5. Differentiate the following from first principles:

(i) sin âˆš2x

(ii) cos âˆšx

Solution:

(i) sin âˆš2x

Given:

f (x) = sin âˆš2x

f (x + h) = sin âˆš2(x+h)

By using the formula,

(ii) cos âˆšx

Given:

f (x) = cos âˆšx

f (x + h) = cos âˆš(x+h)

By using the formula,

#### EXERCISE 30.3 PAGE NO: 30.33

Differentiate the following with respect to x:

1. x4Â â€“ 2sin x + 3 cos x

Solution:

Given:

f (x) = x4Â â€“ 2sin x + 3 cos x

Differentiate on both the sides with respect to x, we get

2. 3xÂ + x3Â + 33

Solution:

Given:

f (x) = 3xÂ + x3Â + 33

Differentiate on both the sides with respect to x, we get

Solution:

Given:

Differentiate on both the sides with respect to x, we get

4. ex log aÂ + ea log xÂ + ea log a

Solution:

Given:

f (x) = ex log aÂ + ea log xÂ + ea log a

We know that,

elog f(x) =Â f(x)

So,

f(x) = axÂ + xaÂ + aa

Differentiate on both the sides with respect to x, we get

5. (2x2Â + 1) (3x + 2)

Solution:

Given:

f (x) = (2x2Â + 1) (3x + 2)

= 6x3Â + 4x2Â + 3x + 2

Differentiate on both the sides with respect to x, we get

#### EXERCISE 30.4 PAGE NO: 30.39

Differentiate the following functions with respect to x:

1. x3Â sin x

Solution:

Let us consider y = x3Â sin x

We need to find dy/dx

We know that y is a product of two functions say u and v where,

u = x3Â and v = sin x

âˆ´Â y = uv

Now let us apply product rule of differentiation.

By using product rule, we get

2. x3Â ex

Solution:

Let us consider y = x3Â ex

We need to find dy/dx

We know that y is a product of two functions say u and v where,

u = x3Â and v = ex

âˆ´Â y = uv

Now let us apply product rule of differentiation.

By using product rule, we get

3. x2Â exÂ log x

Solution:

Let us consider y = x2Â exÂ log x

We need to find dy/dx

We know that y is a product of two functions say u and v where,

u = x2Â and v = ex

âˆ´Â y = uv

Now let us apply product rule of differentiation.

By using product rule, we get

4. xnÂ tan x

Solution:

Let us consider y = xnÂ tan x

We need to find dy/dx

We know that y is a product of two functions say u and v where,

u = xnÂ and v = tan x

âˆ´Â y = uv

Now let us apply product rule of differentiation.

By using product rule, we get

5. xnÂ logaÂ x

Solution:

Let us consider y = xnÂ logaÂ x

We need to find dy/dx

We know that y is a product of two functions say u and v where,

u = xnÂ and v = logaÂ x

âˆ´Â y = uv

Now let us apply product rule of differentiation.

By using product rule, we get

#### EXERCISE 30.5 PAGE NO: 30.44

Differentiate the following functions with respect to x:

Solution:

Let us consider

y =

We need to find dy/dx

We know that y is a fraction of two functions say u and v where,

u = x2 + 1Â and v = x + 1

âˆ´Â y = u/v

Now let us apply quotient rule of differentiation.

By using quotient rule, we get

Solution:

Let us consider

y =

We need to find dy/dx

We know that y is a fraction of two functions say u and v where,

u = 2x – 1Â and v = x2 + 1

âˆ´Â y = u/v

Now let us apply quotient rule of differentiation.

By using quotient rule, we get

Solution:

Let us consider

y =

We need to find dy/dx

We know that y is a fraction of two functions say u and v where,

u = x + exÂ and v = 1 + log x

âˆ´Â y = u/v

Now let us apply quotient rule of differentiation.

By using quotient rule, we get

Solution:

Let us consider

y =

We need to find dy/dx

We know that y is a fraction of two functions say u and v where,

u = ex â€“ tan xÂ and v = cot x – xn

âˆ´Â y = u/v

Now let us apply quotient rule of differentiation.

By using quotient rule, we get

Solution:

Let us consider

y =

We need to find dy/dx

We know that y is a fraction of two functions say u and v where,

u = ax2 + bx + cÂ and v = px2 + qx + r

âˆ´Â y = u/v

Now let us apply quotient rule of differentiation.

By using quotient rule, we get

### Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 30 – Derivatives

Exercise 30.1 Solutions

Exercise 30.2 Solutions

Exercise 30.3 Solutions

Exercise 30.4 Solutions

Exercise 30.5 Solutions

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 30

### What is the use of practising RD Sharma Solutions for Class 11 Maths Chapter 30?

Practising RD Sharma Solutions for Class 11 Maths Chapter 30 provides you with an idea about the sample of questions that will be asked in the board exam, which would help students prepare competently. These solutions cover all topics included in the textbook prescribed by the CBSE board. The solutions formulated are useful resources, which can provide students with all the vital information in the most precise form.

### Whether RD Sharma Solutions for Class 11 Maths Chapter 30 can be viewed online?

The RD Sharma Solutions for Class 11 Maths Chapter 30 can be viewed online and can be downloaded for free. For the ease of learning, the solutions have also been provided in PDF format, so that the students can download them for free and refer to the solutions offline as well. The solutions PDF are solved by a highly experienced faculty team of members having vast experience in the subject at BYJUâ€™S as per the CBSE guidelines and exam patterns.

### Is it necessary to learn all the questions provided in RD Sharma Solutions for Class 11 Maths Chapter 30?

Yes, you must learn all the questions provided in RD Sharma Solutions for Class 11 Maths Chapter 30. Because these questions may appear in board exams as well in class tests. The solutions created by the experts are of high quality and the free download links can be used by the students to access them. By learning these questions students will be ready to write precise and elaborate answers for their upcoming exams more confidently.