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