What's the derivative of #arctan(6^x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Eddie Jun 24, 2016 # y' = ( ln(6) 6^x)/( 1 + 6^{2x} )# Explanation: #y=arctan(6^x)# #tan y=6^x# #sec^2 y \ y' = (6^x)'# #z = 6^x# #ln z = x ln(6)# #1/z z' = ln(6)# #z' = ln(6) 6^x# #\implies sec^2 y \ y' = ln(6) 6^x# # y' = ( ln(6) 6^x)/( sec^2 y)# using #tan^2 +1 = sec^2# # y' = ( ln(6) 6^x)/( 1 + 6^{2x} )# Answer link Related questions What is the derivative of #f(x)=sin^-1(x)# ? What is the derivative of #f(x)=cos^-1(x)# ? What is the derivative of #f(x)=tan^-1(x)# ? What is the derivative of #f(x)=sec^-1(x)# ? What is the derivative of #f(x)=csc^-1(x)# ? What is the derivative of #f(x)=cot^-1(x)# ? What is the derivative of #f(x)=(cos^-1(x))/x# ? What is the derivative of #f(x)=tan^-1(e^x)# ? What is the derivative of #f(x)=cos^-1(x^3)# ? What is the derivative of #f(x)=ln(sin^-1(x))# ? See all questions in Differentiating Inverse Trigonometric Functions Impact of this question 1575 views around the world You can reuse this answer Creative Commons License