Answer
$$\frac{d}{{dx}}\left( {{x^\pi } + {\pi ^x}} \right) = \pi {x^{\pi - 1}} + {\pi ^x}\ln \pi $$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {{x^\pi } + {\pi ^x}} \right) \cr
& {\text{sum rule for differentiation}} \cr
& \frac{d}{{dx}}\left( {{x^\pi } + {\pi ^x}} \right) = \frac{d}{{dx}}\left( {{x^\pi }} \right) + \frac{d}{{dx}}\left( {{\pi ^x}} \right) \cr
& {\text{compute derivatives}} \cr
& \frac{d}{{dx}}\left( {{x^\pi } + {\pi ^x}} \right) = \pi {x^{\pi - 1}} + {\pi ^x}\left( {\ln \pi } \right)\left( 1 \right) \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}\left( {{x^\pi } + {\pi ^x}} \right) = \pi {x^{\pi - 1}} + {\pi ^x}\ln \pi \cr} $$
