#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 29 sub question (i) maths textbook solution

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

Hint: To solve this equation we use $f(x)^{f(x)}$  formula

Given:

Solution:

$\begin{gathered} y=x^{\cos x}+\sin x^{\tan x} \\\\ y=(f(x))^{f(x)} \end{gathered}$

\begin{aligned} &\text { Let } u=x^{\cos x} \\\\ &\log u=x^{\cos x} \\\\ &\log u=\cos x \log x \end{aligned}

\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=\cos x \cdot \frac{1}{x}+\log x(-\sin x) \\\\ &\frac{d u}{d x}=u\left(\frac{\cos x}{x}-\sin x \log x\right) \end{aligned}

\begin{aligned} &=x \cos x\left[\left(\frac{\cos x}{x}-\sin x \log x\right)\right] \\\\ &v=\sin x^{\tan x} \\\\ &\log v=\tan x \log \sin x \end{aligned}

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

$\frac{d v}{d x}=v\left(\tan x \times \frac{\cos x}{\sin x}+\sec ^{2} x \log \sin x\right)$

$\frac{d v}{d x}=v\left(\frac{\sin x}{\cos x} \times \frac{\cos x}{\sin x}+\sec \sin ^{2} x \log \sin x\right)$

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

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$\frac{d v}{d x}=x \cos x\left[\left(\frac{\cos x}{x}-\sin x \log x\right)\right]+\sin x^{\tan x}\left ( 1+\sec ^{2} x\log \sin x\right )$