If y = x ^{sin x} + (sin x) ^x , then $\frac{dy}{dx}$ =

The given function is $y={\left(x\right)}^{\sin x}+{\left(\sin x\right)}^x$

Suppose that $f\left(x\right)=x^{\sin x}$ and $g\left(x\right)={\left(\sin x\right)}^x$

Use the logartithmic differentiation to differentiate both functions

$\log \left(f\left(x\right)\right)=\sin x·\log x$

differentiate both  sides

$\frac{f\text{'}\left(x\right)}{f\left(x\right)}=\sin x·\frac{1}{x}+\log x\left(\cos x\right)$

Hence, $f\text{'}\left(x\right)=x^{\sin x}\left(\sin x·\frac{1}{x}+\log x\left(\cos x\right)\right)$

Similarly the other function

$\log \left(g\left(x\right)\right)=x\log \left(\sin x\right)$

differentiate both sides

$\frac{1}{g\left(x\right)}g\text{'}\left(x\right)=x·\frac{1}{\sin x}\cos x+\log \left(\sin x\right)$

Hence, 

$g\text{'}\left(x\right)={\left(\sin x\right)}^x\left(x cotx+\log \left(\sin x\right)\right)$

Therefore, 

  $$\begin{gathered} \frac{dy}{dx}=f\text{'}\left(x\right)+g\text{'}\left(x\right) \\ =x^{\sin x}\left(\sin x·\frac{1}{x}+\log x\left(\cos x\right)\right)+{\left(\sin x\right)}^x\left(x cotx+\log \left(\sin x\right)\right) \end{gathered}$$