How do you differentiate #f(x)= x^2*tan^-1 x#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Bill K. Jun 10, 2015 The answer is #f'(x)=x^2/(1+x^2)+2x tan^{-1}(x)# Explanation: This follows from the Product Rule #d/dx(g(x)h(x))=g(x)h'(x)+g'(x)h(x)# (with #g(x)=x^2# and #h(x)=tan^{-1}(x)#), the Power Rule #d/dx(x^(n))=nx^(n-1)#, and the derivative of inverse tangent #d/dx(tan^{-1}(x))=1/(1+x^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 2355 views around the world You can reuse this answer Creative Commons License