Answer
$$y' = {\pi ^{\sin x}}\cos x\left( {\ln \pi } \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\pi ^{\sin x}} \cr
& y = {\pi ^{\sin x}} \cr
& {\text{take logarithm natural on both sides}} \cr
& \ln y = \ln {\pi ^{\sin x}} \cr
& {\text{logarithm properties}} \cr
& \ln y = \sin x\ln \left( \pi \right) \cr
& {\text{differentiate}} \cr
& \left( {\ln y} \right)' = \left( {\sin x\ln \left( \pi \right)} \right)' \cr
& \left( {\ln y} \right)' = \ln \pi \left( {\sin x} \right)' \cr
& \frac{{y'}}{y} = \ln \pi \left( {\cos x} \right) \cr
& y' = \ln \pi y\cos x \cr
& {\text{replace }}y = {\pi ^{\sin x}} \cr
& y' = \ln \pi \left( {{\pi ^{\sin x}}} \right)\cos x \cr
& {\text{simplify}} \cr
& y' = {\pi ^{\sin x}}\cos x\left( {\ln \pi } \right) \cr} $$