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  • Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 19

Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 19

Answers (1)

Answer:

\cos \operatorname{ec} 2 x

Hint:

By using the chain rule of differentiation

Given:

y=\log \sqrt{\tan x}
Solution:  

\text { i. } e \cdot \frac{d y}{d x}=\frac{d(\log \sqrt{\tan x})}{d(\sqrt{\tan x})} \cdot \frac{d(\sqrt{\tan x})}{d(\tan x)} \cdot \frac{d(\tan x)}{d x}

\begin{aligned} &=\frac{1}{\sqrt{\tan x}} \cdot \frac{1}{2 \sqrt{\tan x}} \cdot \sec ^{2} x \\\\ &=\frac{\sec ^{2} x}{2 \tan x} \\\\ &=\frac{1+\tan ^{2} x}{2 \tan x} \end{aligned}

\begin{aligned} &=\frac{1}{2}(\tan x+\cot x) \\\\ &=\frac{1}{2}\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right] \end{aligned}

=\frac{1}{2}\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right] \quad\left[\because \sin ^{2} x+\cos ^{2} x=1\right]

=\frac{1}{2} \times \frac{2}{\sin 2 x}=\cos e c 2 x \quad[\sin 2 x=2 \sin x \cos x]

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