Differentiate the following w.r.t. x:
${\left(x\cos x\right)}^x+{\left(x\sin x\right)}^{\frac{1}{x}}$

Solution: Let $y={\left(x\cos x\right)}^x+{\left(x\sin x\right)}^{\frac{1}{x}}$
Let $u={\left(x\cos x\right)}^x\text{and}v={\left(x\sin x\right)}^{\frac{1}{x}}$
Then, $y=u+v$
$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Now, $u={\left(x\cos x\right)}^x$
Taking log both sides,
$\log u=x\left(\log x+\log \cos x\right)$
Differentiating both sides,
$\frac{1}{u}\frac{du}{dx}=\left(\log x+\cos x\right)\cdot 1+x\left(\frac{1}{x}+\frac{1}{\cos x}\left(-\sin x\right)\right)$
$\frac{1}{u}\frac{du}{dx}=\left(\log x+\cos x\right)+1-x\tan x$
$\frac{du}{dx}={\left(x\cos x\right)}^x\left(\log \left(x\cos x\right)+1-x\tan x\right)\to \left(1\right)$
Now, $v={\left(x\sin x\right)}^{\frac{1}{x}}$
Taking log both sides,
$\log v=\frac{1}{x}\left(\log x+\log \sin x\right)$
Differentiating both sides,
$\frac{1}{v}\frac{dv}{dx}=\frac{1}{x^2}\left(x\left(\frac{1}{x}+\frac{1}{\sin x}\left(\cos x\right)\right)-\left(\log x+\log \sin x\right)\cdot 1\right)$
$\frac{1}{v}\frac{dv}{dx}=\frac{1}{x^2}\left(1+\cot x-\log \left(x\sin x\right)\right)$
$\frac{dv}{dx}={\left(x\sin x\right)}^{\frac{1}{x}}\frac{1}{x^2}\left(1+\cot x-\log \left(x\sin x\right)\right)$
So, $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\frac{dy}{dx}={\left(x\cos x\right)}^x\left(\log \left(x\cos x\right)+1-x\tan x\right)+{\left(x\sin x\right)}^{\frac{1}{x}}\left(\frac{1+\cot x-\log \left(x\sin x\right)}{x^2}\right)$