#### Answer

$${x^{\tan x}}\left( {\frac{{\tan x}}{x} + {{\sec }^2}x\ln x} \right)$$

#### Work Step by Step

$$\eqalign{
& \frac{d}{{dx}}\left( {{x^{\tan x}}} \right) \cr
& {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr
& = {e^{\ln {x^{\tan x}}}} \cr
& = {e^{\tan x\ln x}} \cr
& {\text{evaluate the derivative}} \cr
& = \frac{d}{{dx}}\left( {{e^{\tan x\ln x}}} \right) \cr
& {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr
& = {e^{\tan x\ln x}}\frac{d}{{dx}}\left( {\tan x\ln x} \right) \cr
& {\text{product rule}} \cr
& = {e^{\tan x\ln x}}\left( {\frac{{\tan x}}{x} + {{\sec }^2}x\ln x} \right) \cr
& {\text{simplify}} \cr
& = {x^{\tan x}}\left( {\frac{{\tan x}}{x} + {{\sec }^2}x\ln x} \right) \cr} $$