Solution:
Let’s take $y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{2}}$
Putting the value of $u=(x \cos x)^{x}$ and ${ }^{v=(x \sin x)^{\frac{1}{x}}}$,
we obtain $y=u+v$
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \dots \dots \dots$(1)
So now $u=(x \cos x)^{x}$
$\log u=\log (x \cos x)^{x}=x \log (x \cos x)$
$\log u=x(\log x+\log \cos x)$
$\begin{aligned}
&\frac{d}{d x} \log u=\frac{d}{d x}\{x(\log x+\log \cos x)\} \\
&\frac{1}{u} \frac{d u}{d x}=x\left[\frac{1}{x}+\frac{1}{\cos x} \cdot(-\sin x)\right]+(\log x+\log \cos x) \cdot 1
\end{aligned}$
$\begin{aligned}
&\frac{d u}{d x}=u[1-x \tan x+\log (x \cos x)] \\
&\frac{d u}{d x}=(x \cos x)^{x}[1-x \tan x+\log (x \cos x)]
\end{aligned}$
$\begin{aligned}
&\text {Now again } v=(x \sin x)^{\frac{1}{x}} \\
&\log v=\log (x \sin x)^{\frac{1}{x}}=\frac{1}{x} \log (x \sin x) \\
&\log v=\frac{1}{x}(\log x+\log \sin x) \\
&\frac{d}{d x} \log v=\frac{d}{d x}\left\{\frac{1}{x}(\log x+\log \sin x)\right\}
\end{aligned}$
$\begin{aligned}
&\frac{1}{v} \frac{d v}{d x}=\frac{1}{x}\left[\frac{1}{x}+\frac{1}{\sin x} \cdot \cos x\right]+(\log x+\log \sin x)\left(\frac{-1}{x^{2}}\right) \\
&\frac{d v}{d x}=v\left[\frac{1}{x^{2}}+\frac{\cot x}{x}-\frac{\log (x \sin x)}{x^{2}}\right] \\
&\frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1}
{x^{2}}+\frac{\cot x}{x}-\frac{\log (x \sin x)}{x^{2}}\right]_{\dots \dots \dots \dots (3)}\end{aligned}$
Putting the values from equations (2) and (3) in equation (1)
$\frac{d y}{d x}=(x \cos x)^{x}[1-x \tan x+\log (x \cos x)]+(x \sin x)^{\frac{1}{x}}\left[\frac{1}{x^{2}}+\frac{\cot x}{x}-\frac{\log (x \sin x)}{x^{2}}\right]$