What's the derivative of #arctan(x)/(1+x^2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Andrea S. Oct 6, 2017 #d/dx ( arctanx/(1+x^2) ) = ( 1- 2xarctanx )/(1+x^2)^2# Explanation: Using the quotient rule: #d/dx (f/g) = (g * (df)/dx - f * (dg)/dx)/g^2# we have: #d/dx ( arctanx/(1+x^2) ) = ( (1+x^2) d/dx arctanx - arctanx d/dx (1+x^2))/(1+x^2)^2# #d/dx ( arctanx/(1+x^2) ) = ( (1+x^2)(1/(1+x^2)) -2x arctanx )/(1+x^2)^2# #d/dx ( arctanx/(1+x^2) ) = ( 1- 2xarctanx )/(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 2507 views around the world You can reuse this answer Creative Commons License