How do you differentiate # f(x) = arctan(1/(1+x^2)) #? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Bdub Mar 10, 2016 #f'(x) = 1/(1+(1/(1+x^2))^2 ) *-(2x)/(1+x^2)^2# Explanation: #f'(u)=1/(1+u^2) * u'# #f'(x) = 1/(1+(1/(1+x^2))^2 ) *-1/(1+x^2)^2 (2x)# #f'(x) = 1/(1+(1/(1+x^2))^2 ) -(2x)/(1+x^2)^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 3123 views around the world You can reuse this answer Creative Commons License