Answer
$$y'=(\ln \sin x+x\cot x)(\sin x)^x$$
Work Step by Step
$$y=(\sin x)^x$$
We can also use logarithmic differentiation to solve these types of problems. Take the natural logarithm of both sides: $$\ln y=\ln(\sin x)^x$$ $$\ln y =x\ln\sin x$$
Now use implicit differentiation to find $y'$ with respect to $x$: $$(\ln y)' =(x\ln\sin x)'$$
$$\frac{y'}{y}=\ln \sin x+x\times\frac{1}{\sin x}\times(\sin x)'$$
$$\frac{y'}{y}=\ln \sin x+\frac{x\cos x}{\sin x}=\ln \sin x+x\cot x$$
Find $y'$: $$y'=(\ln \sin x+x\cot x)y=(\ln \sin x+x\cot x)(\sin x)^x$$