If $y=x^x-2^{\sin x}$, then $\frac{dy}{dx}=$

$y=x^x-2^{\sin x}$

$y=u-v$

$\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$

$u=x^x$

$\log u=x\log x$

$\frac{1}{u}\frac{du}{dx}=\frac{x}{x}+\log x$

$\frac{du}{dx}=x^x(1+\log x)$

$v=2^{\sin x}$

$\log v=\sin x\log 2$

$\frac{1}{v}\frac{dv}{dx}=\log 2\left(\cos x\right)$

$\frac{dv}{dx}=2^{\sin x}\left(\cos x\log 2\right)$

$\frac{dy}{dx}=x^x(1+\log x)-2^{\sin x}\left(\cos x\log 2\right)$