Given function:
#f(x)=e^{-x}\cos(-x)-\frac{\sin x}{\pi-e^x}#
#f(x)=e^{-x}\cosx+\frac{\sin x}{e^x-\pi}#
Differentiating above function w.r.t. #x# using chain rule & quotient rule as follows
#f'(x)=d/dx(e^{-x}\cosx+\frac{\sin x}{e^x-\pi})#
#=e^{-x}(-\sinx)+\cos x(-e^{-x})+\frac{(e^x-\pi)\cos x-\sin x(e^x)}{(e^x-\pi)^2}#
#=-e^{-x}(\sinx+\cos x)+\frac{e^x(\cos x-\sin x)-\pi\cos x}{(e^x-\pi)^2}#
Setting #x=\pi/6# in above expression we get
#f'(\pi/6)#
#=-e^{-\pi/6}(\sin(\pi/6)+\cos (\pi/6))+\frac{e^{\pi/6}(\cos (\pi/6)-\sin (\pi/6))-\pi\cos (\pi/6)}{(e^{\pi/6}-\pi)^2}#
#=-0.80921-0.9953#
#=-1.8045#
Since #f'(\pi/6)<0# hence the given function #f(x)# is decreasing at #x=\pi/6#
