What is the derivative of #arctan (8/x^2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer moutar Apr 1, 2016 #(-16x)/(x^4+64)# Explanation: By the chain rule: #y=arctanu, u = 8/x^2# #dy/(du) = 1/(1+u^2), (du)/dx = -16/x^3# #dy/dx=dy/(du)*(du)/dx # #= 1/(1+(8/x^2)^2)*-16/x^3# #= -16/(x^3+64/x) = (-16x)/(x^4+64)# 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 1588 views around the world You can reuse this answer Creative Commons License