If #f(x)= x^2*tan^-1 x# then how do you find f'(1)? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Konstantinos Michailidis Sep 13, 2015 It is #f'(1)=1/2-pi/2# Explanation: Since #f(x)=x^2*tan^-1(x)# the derivative is #f'(x)=2x*tan^-1(x)+x^2*1/(1+x^2)# hence #f'(1)=2*tan^(-1)(-1)+1/2# But #tan^(-1)(-1)=tan^(-1)(tan(-pi/4))=-pi/4# So #f'(1)=1/2-pi/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 5046 views around the world You can reuse this answer Creative Commons License