Solve this

let $y=x^{\cos x}+(\sin x)^{\tan x}$

$\Rightarrow y=a+b$

where $a=x^{\cos x} ; b=(\sin x)^{\tan x}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where $\mathrm{a}$ and $\mathrm{u}$ are any variables $\}$

$a=x^{\cos x}$

Taking log both the sides:

$\Rightarrow \log a=\log (x)^{\cos x}$

$\Rightarrow \log a=\cos x \log x$

$\left\{\log x^{a}=a \log x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{d(\log a)}{d x}=\frac{d(\cos x \log x)}{d x}$

$\Rightarrow \frac{d(\log a)}{d x}=\cos x \times \frac{d(\log x)}{d x}+\log x \times \frac{d(\cos x)}{d x}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\cos x \times \frac{1}{x} \frac{d x}{d x}+\log x(-\sin x)$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\frac{\cos x}{x}-\sin x \log x$

$\Rightarrow \frac{d a}{d x}=a\left\{\frac{\cos x}{x}-\sin x \log x\right\}$

Put the value of $a=x^{\cos x}$ :

$\Rightarrow \frac{d a}{d x}=x^{\cos x}\left\{\frac{\cos x}{x}-\sin x \log x\right\}$

$b=(\sin x)^{\tan x}$

Taking log both the sides:

$\Rightarrow \log b=\log (\sin x)^{\tan x}$

$\Rightarrow \log b=\tan x \log (\sin x)$

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\frac{\mathrm{d}(\tan \mathrm{x} \log (\sin \mathrm{x}))}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\tan \mathrm{x} \times \frac{\mathrm{d}(\log (\sin \mathrm{x}))}{\mathrm{dx}}+\log (\sin \mathrm{x}) \times \frac{\mathrm{d}(\tan \mathrm{x})}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

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

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} ; \frac{d(\tan x)}{d x}=\sec ^{2} x ; \frac{d(\sin x)}{d x}=\cos x\right\}$

$\Rightarrow \frac{1}{\mathrm{~b}} \frac{\mathrm{db}}{\mathrm{dx}}=\frac{\sin \mathrm{x}}{\cos \mathrm{x}} \times \frac{1}{\sin \mathrm{x}}(\cos \mathrm{x})+\sec ^{2} \mathrm{x} \log (\sin \mathrm{x})$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\mathrm{b}\left\{1+\sec ^{2} \mathrm{x} \log (\sin \mathrm{x})\right\}$

Put the value of $b=(\sin x)^{\tan x}$ :

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=(\sin \mathrm{x})^{\tan x}\left\{1+\sec ^{2} x \log (\sin x)\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

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