How do you get the derivative of #(arctan(6x^2 +5))^2 #? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Massimiliano May 25, 2015 In this way with the chain rule: #y'=2arctan(6x^2+5)*1/(1+(6x^2+5)^2)*12x#. 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 1486 views around the world You can reuse this answer Creative Commons License