Answer
$$\eqalign{
& f'\left( x \right) = {\left( {\sin x} \right)^{\ln x}}\left[ {\frac{{x\ln x\cot x + \sin x\ln \left( {\sin x} \right)}}{x}} \right] \cr
& f'\left( {\frac{\pi }{2}} \right) = 0 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {\sin x} \right)^{\ln x}};\,\,\,\,\,\,a = \frac{\pi }{2} \cr
& {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& f\left( x \right) = {e^{\ln x\ln \left( {\sin x} \right)}} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = {e^{\ln x\ln \left( {\sin x} \right)}}\frac{d}{{dx}}\left[ {\ln x\ln \left( {\sin x} \right)} \right] \cr
& f'\left( x \right) = {\left( {\sin x} \right)^{\ln x}}\left[ {\ln x\left( {\frac{{\cos x}}{{\sin x}}} \right) + \frac{1}{x}\ln \left( {\sin x} \right)} \right] \cr
& f'\left( x \right) = {\left( {\sin x} \right)^{\ln x}}\left[ {\frac{{x\ln x\cot x + \sin x\ln \left( {\sin x} \right)}}{x}} \right] \cr
& \cr
& {\text{Evaluate the derivative at }}x = \frac{\pi }{2} \cr
& f'\left( {\frac{\pi }{2}} \right) = {\left( {\sin \frac{\pi }{2}} \right)^{\ln \frac{\pi }{2}}}\left[ {\frac{{\frac{\pi }{2}\ln \frac{\pi }{2}\cot \frac{\pi }{2} + \sin \frac{\pi }{2}\ln \left( {\sin \frac{\pi }{2}} \right)}}{{\pi /2}}} \right] \cr
& f'\left( {\frac{\pi }{2}} \right) = {\left( 1 \right)^{\ln \frac{\pi }{2}}}\left[ {\frac{{0 + 0}}{{\frac{\pi }{2}}}} \right] \cr
& f'\left( {\frac{\pi }{2}} \right) = 0 \cr} $$
