How do you differentiate #y=(sin^-1x)/(1+x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Alan N. Dec 12, 2016 #dy/dx=1/((1+x)sqrt(1-x^2)) - sin^-1x/(1+x)^2# Explanation: #y=sin^-1x/(1+x)# Apply the quotient rule #dy/dx= ((1+x)* d/dx sin^-1x - sin^-1x*d/dx(1+x))/(1+x)^2# #= ((1+x)* 1/sqrt(1+x^2) - sin^-1x* 1)/(1+x)^2# (Standard differential) #= 1/((1+x)sqrt(1-x^2)) - sin^-1x/(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 2974 views around the world You can reuse this answer Creative Commons License