## Calculus, 10th Edition (Anton)

$$y' = \frac{{\tan x{{\left( {\ln x} \right)}^{\tan x - 1}}}}{x} + {\left( {\ln x} \right)^{\tan x}}\ln \left( {\ln x} \right){\sec ^2}x$$
\eqalign{ & y = {\left( {\ln x} \right)^{\tan x}} \cr & {\text{take logarithm natural on both sides}} \cr & \ln y = \ln {\left( {\ln x} \right)^{\tan x}} \cr & {\text{logarithm properties}} \cr & \ln y = \tan x\ln \left( {\ln x} \right) \cr & {\text{differentiate}} \cr & \left( {\ln y} \right)' = \left( {\tan x\ln \left( {\ln x} \right)} \right)' \cr & {\text{product rule}} \cr & \frac{{y'}}{y} = \tan x\left( {\ln \left( {\ln x} \right)} \right)' + \ln \left( {\ln x} \right)\left( {\tan x} \right)' \cr & \frac{{y'}}{y} = \tan x\left( {\frac{{\frac{1}{x}}}{{\ln x}}} \right) + \ln \left( {\ln x} \right)\left( {{{\sec }^2}x} \right) \cr & \frac{{y'}}{y} = \tan x\left( {\frac{1}{{x\ln x}}} \right) + \ln \left( {\ln x} \right){\sec ^2}x \cr & {\text{solve for }}y' \cr & y' = y\left( {\tan x\left( {\frac{1}{{x\ln x}}} \right) + \ln \left( {\ln x} \right){{\sec }^2}x} \right) \cr & {\text{replace }}y = {\left( {\ln x} \right)^{\tan x}} \cr & y' = {\left( {\ln x} \right)^{\tan x}}\left( {\tan x\left( {\frac{1}{{x\ln x}}} \right)' + \ln \left( {\ln x} \right){{\sec }^2}x} \right) \cr & {\text{simplify}} \cr & y' = \frac{{\tan x{{\left( {\ln x} \right)}^{\tan x - 1}}}}{x} + {\left( {\ln x} \right)^{\tan x}}\ln \left( {\ln x} \right){\sec ^2}x \cr}