Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(e) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(e)

Question 1.

Find derivatives of the following functions from the definition :

(i) 3x^{2} – \(\frac{4}{x}\)

Solution:

(ii) (4x – 1)^{2}

Solution:

(iii) 2 + x + √x^{3}

Solution:

(iv) x – \(\sqrt{x^2-1}\)

Solution:

(v) \(\frac{1}{x^{2 / 5}}\) + 1

Solution:

Question 2.

(i) cos (ax + b)

Solution:

Let y = cos (ax + b)

Then \(\frac{d y}{d x}\) = -sin (ax + b) × \(\frac{d}{d x}\) (ax + b) by chain rule.

= -sin(ax + b). a = -a sin (ax + b)

(ii) x^{2} sin x

Solution:

Let y = x^{2} sin x

Then \(\frac{d y}{d x}=\frac{d}{d x}\) (x^{2}). sin x + x^{2} \(\frac{d}{d x}\)

[ ∴ \(\frac{\mathrm{d}}{\mathrm{dx}}(u \cdot v)=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x}\)

= 2x sin x + x^{2} cos x

(iii) \(\sqrt{\tan x}\)

Solution:

Ley y = \(\sqrt{\tan x}\) = \((\tan x)^{\frac{1}{2}}\)

Then \(\frac{d y}{d x}=\frac{1}{2}(\tan x)^{-\frac{1}{2}} \times \frac{d}{d x}\)(tan x)

= \(\frac{1}{2 \sqrt{\tan x}}\) sec^{2} x.

(iv) cot x^{2}

Solution:

Let y = cot x^{2}

Then \(\frac{d y}{d x}=-{cosec}^2 x^2 \times \frac{d}{d x}\left(x^2\right)\)

= – cosec^{2} x^{2}. 2x

= -2x. cosec^{2} x^{2}

(v) cosec 3x

Solution:

Let y = cosec 3x

Then \(\frac{d y}{d x}\) = -3 cosec 3x . cot 3x

Question 3.

(i) √x sin x

Solution:

Let y = √x sin x

Then \(\frac{d y}{d x}=\frac{d}{d x}\)(√x) sin x + √x. \(\frac{d}{d x}\)(sin x)

= \(\frac{1}{2 \sqrt{x}}\) sin x + √x. cos x

(ii) \(\sqrt{x^2+1}\)cos x

Solution:

(iii) tan x – x^{2} – 2x

Solution:

Let y = tan x – x^{2} – 2x

\(\frac{d y}{d x}\) = sec^{2} x – 2x – 2