Answer
$$y' = {\pi ^{x\tan x}}\ln \pi \left( {x{{\sec }^2}x + \tan x} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\pi ^{x\tan x}} \cr
& y = {\pi ^{x\tan x}} \cr
& {\text{take logarithm natural on both sides}} \cr
& \ln y = \ln {\pi ^{x\tan x}} \cr
& {\text{logarithm properties}} \cr
& \ln y = x\tan x\ln \left( \pi \right) \cr
& {\text{differentiate}} \cr
& \left( {\ln y} \right)' = \left( {x\tan x\ln \left( \pi \right)} \right)' \cr
& \left( {\ln y} \right)' = \ln \pi \left( {x\tan x} \right)' \cr
& {\text{product rule}} \cr
& \frac{{y'}}{y} = \ln \pi \left( {x{{\sec }^2}x + \tan x} \right) \cr
& y' = y\ln \pi \left( {x{{\sec }^2}x + \tan x} \right) \cr
& {\text{replace }}y = {\pi ^{x\tan x}} \cr
& y' = {\pi ^{x\tan x}}\ln \pi \left( {x{{\sec }^2}x + \tan x} \right) \cr} $$
