Answer
$$f'\left( x \right) = {\left( {\sin x} \right)^{\tan x}}\left[ {1 + \left( {{{\sec }^2}x} \right)\ln \left( {\sin x} \right)} \right]$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {\sin x} \right)^{\tan x}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln {\left( {\sin x} \right)^{\tan x}} \cr
& {\text{power property for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = \tan x\ln \left( {\sin x} \right) \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln \left( {f\left( x \right)} \right)} \right] = \frac{d}{{dx}}\left[ {\tan x\ln \left( {\sin x} \right)} \right] \cr
& {\text{use the product rule for differentiation}} \cr
& \frac{d}{{dx}}\left[ {\ln \left( {f\left( x \right)} \right)} \right] = \tan x\frac{d}{{dx}}\left[ {\ln \left( {\sin x} \right)} \right] + \ln \left( {\sin x} \right)\frac{d}{{dx}}\left[ {\tan x} \right] \cr
& {\text{computing derivatives}} \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \tan x\left( {\frac{{\cos x}}{{\sin x}}} \right) + \ln \left( {\sin x} \right)\left( {{{\sec }^2}x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 1 + \ln \left( {\sin x} \right)\left( {{{\sec }^2}x} \right) \cr
& {\text{solving for }}f'\left( x \right) \cr
& f'\left( x \right) = f\left( x \right)\left[ {1 + \left( {{{\sec }^2}x} \right)\ln \left( {\sin x} \right)} \right] \cr
& {\text{replace }}f\left( x \right){\text{ with the original function:}} \cr
& f'\left( x \right) = {\left( {\sin x} \right)^{\tan x}}\left[ {1 + \left( {{{\sec }^2}x} \right)\ln \left( {\sin x} \right)} \right] \cr} $$
