What is the derivative of #arctan(y/x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Joel Kindiak Aug 10, 2015 #- y/(x^2 + y^2) dy/dx# Explanation: #d/(dx) arctan(y/x) = 1/(1 + (y/x)^2) * (d/dx y/x)# #d/(dx) arctan(y/x) = 1/(1 + (y^2/x^2)) * (- y/x^2 * dy/dx)# #d/(dx) arctan(y/x) = - x^2/(x^2 + y^2) * y/x^2 * dy/dx# #d/(dx) arctan(y/x) = - y/(x^2 + y^2) dy/dx# 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 50216 views around the world You can reuse this answer Creative Commons License