Answer
$$\eqalign{
& f'\left( x \right) = {\left( {\tan x} \right)^{x - 1}}\left[ {\frac{{x - 1 + \left( {\sin x\cos x} \right)\ln \left( {\tan x} \right)}}{{\sin x\cos x}}} \right] \cr
& f'\left( {\frac{\pi }{4}} \right) = \frac{\pi }{2} - 2 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {\tan x} \right)^{x - 1}};{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} a = \frac{\pi }{4} \cr
& {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& f\left( x \right) = {e^{\left( {x - 1} \right)\ln \left( {\tan x} \right)}} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = {e^{\left( {x - 1} \right)\ln \left( {\tan x} \right)}}\frac{d}{{dx}}\left[ {\left( {x - 1} \right)\ln \left( {\tan x} \right)} \right] \cr
& f'\left( x \right) = {\left( {\tan x} \right)^{x - 1}}\left[ {\left( {x - 1} \right)\left( {\frac{1}{{\sin x\cos x}}} \right) + \ln \left( {\tan x} \right)} \right] \cr
& f'\left( x \right) = {\left( {\tan x} \right)^{x - 1}}\left[ {\frac{{x - 1}}{{\sin x\cos x}} + \ln \left( {\tan x} \right)} \right] \cr
& f'\left( x \right) = {\left( {\tan x} \right)^{x - 1}}\left[ {\frac{{x - 1 + \left( {\sin x\cos x} \right)\ln \left( {\tan x} \right)}}{{\sin x\cos x}}} \right] \cr
& {\text{Evaluate the derivative at }}x = \frac{\pi }{4} \cr
& f'\left( {\frac{\pi }{4}} \right) = {\left( {\tan \frac{\pi }{4}} \right)^{\frac{\pi }{4} - 1}}\left[ {\frac{{\frac{\pi }{4} - 1 + \left( {\sin \frac{\pi }{4}\cos \frac{\pi }{4}} \right)\ln \left( {\tan \frac{\pi }{4}} \right)}}{{\sin \frac{\pi }{4}\cos \frac{\pi }{4}}}} \right] \cr
& {\text{Simplifying}} \cr
& f'\left( {\frac{\pi }{4}} \right) = {\left( 1 \right)^{\frac{\pi }{4} - 1}}\left[ {\frac{{\frac{\pi }{4} - 1 + \left( {1/2} \right)\ln \left( 1 \right)}}{{1/2}}} \right] \cr
& f'\left( {\frac{\pi }{4}} \right) = \left[ {\frac{{\frac{\pi }{4} - 1 + 0}}{{1/2}}} \right] \cr
& f'\left( {\frac{\pi }{4}} \right) = \frac{\pi }{2} - 2 \cr} $$
