#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 30

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

Hint: Differentiate the equation taking log on both sides

Given:   $y=(\tan x)^{\log x}+\cos ^{2}\left(\frac{\pi}{4}\right)$

Solution:

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(u+\cos ^{2}\left(\frac{\pi}{4}\right)\right) \\\\ &\frac{d y}{d x}=\frac{d u}{d x} \end{aligned}

Let $u=(\tan x)^{\log x}$

\begin{aligned} &\log u=\log (\tan x)^{\log x} \\\\ &\frac{d}{d x}(\log u)=\log (\tan x)^{\log x} \end{aligned}

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

$\frac{d u}{d x}=u\left[\frac{\log (\tan x)}{x}+\log x \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\right]$

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