How do you differentiate #arctan(x/a)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Eddie Aug 20, 2016 #= a/ (x^2 + a^2)# Explanation: #y = arctan (x/a)# #tan y = x/a# #sec^2 y \ y' = 1/a# # y' = 1/(a sec^2 y)# #= 1/(a (tan^2 y + 1)# #= 1/(a ((x/a)^2 + 1)# #= a/ (x^2 + a^2)# 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 1204 views around the world You can reuse this answer Creative Commons License