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#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 60 maths

Answer: $\frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{x}{\sqrt{2\left(x^{2}+1\right)}}$

Hint: To solve this equation we use log on both side

Given: $y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$

Solution:  $\frac{d}{d x}(\log u)=\log (x \tan x)=\tan x-\log x$

$\text { Let } \mathrm{u}=x^{\tan x}$

$\frac{1}{u} \frac{d u}{d x}=\tan x \frac{d}{d x} \log x+\log x \cdot \frac{d}{d x}(\tan x)$

\begin{aligned} &\frac{d u}{d x}=u\left(\frac{\tan x}{x}+\left(\log x \cdot \sec ^{2} x\right)\right) \\\\ &\frac{d u}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\left(\log x \cdot \sec ^{2} x\right)\right) \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(u+\sqrt{\frac{x^{2}+1}{2}}\right) \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d}{d x}\left(\sqrt{\frac{x^{2}+1}{2}}\right) \end{aligned}

$\frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x\right)$  $+\frac{1}{2 \sqrt{\frac{x^{2}+1}{2}}} \frac{d}{d x}\left(\frac{x^{2}+1}{2}\right)$                \begin{aligned} &{\left[\because \frac{d}{d x} \sqrt{x}=\frac{1}{2 \sqrt{x}}\right]} \\ &{\left[\because \frac{d}{d x} x^{2}=2 x\right]} \end{aligned}

$=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{1}{\sqrt{2\left(x^{2}+1\right)}} \cdot \frac{2 x}{2}$

$\frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{x}{\sqrt{2\left(x^{2}+1\right)}}$