Answer
$$y' = {x^{\sin x - 1}}\sin x + {x^{\sin x}}\ln x\left( {\cos x} \right)$$
Work Step by Step
$$\eqalign{
& y = {x^{\sin x}} \cr
& {\text{take logarithm natural on both sides}} \cr
& \ln y = \ln {x^{\sin x}} \cr
& {\text{logarithm properties}} \cr
& \ln y = \sin x\ln x \cr
& {\text{differentiate}} \cr
& \left( {\ln y} \right)' = \left( {\sin x\ln x} \right)' \cr
& {\text{product rule}} \cr
& \frac{{y'}}{y} = \sin x\left( {\ln x} \right)' + \ln x\left( {\sin x} \right)' \cr
& \frac{{y'}}{y} = \sin x\left( {\frac{1}{x}} \right) + \ln x\left( {\cos x} \right) \cr
& y' = y\left[ {\frac{{\sin x}}{x} + \ln x\left( {\cos x} \right)} \right] \cr
& {\text{replace }}y = {x^{\sin x}} \cr
& y' = {x^{\sin x}}\left[ {\frac{{\sin x}}{x} + \ln x\left( {\cos x} \right)} \right] \cr
& {\text{simplify}} \cr
& y' = {x^{\sin x - 1}}\sin x + {x^{\sin x}}\ln x\left( {\cos x} \right) \cr} $$