What is the derivative of #arcsin(x^2/4)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Bill K. Oct 11, 2015 #(2x)/sqrt(16-x^4)# for #-2 < x < 2# Explanation: Use the Chain Rule, along with the fact that #d/dx(arcsin(x))=1/sqrt(1-x^2)# to get: #d/dx(arcsin(x^2/4))=1/sqrt(1-(x^2/4)^2) * d/dx(x^2/4)# #=1/sqrt(1-x^4/16) * x/2=1/sqrt(1-x^4/16) * (2x)/sqrt(16)# #=(2x)/sqrt(16-x^4)# for #-2 < 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 1386 views around the world You can reuse this answer Creative Commons License