How do you find the derivative of #y=csc^-1 (x/2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Shwetank Mauria Jan 3, 2018 #(dy)/(dx)=-2/(xsqrt(x^2-4))# Explanation: As #y=csc^(-1)(x/2)# #x/2=cscy# taking derivative on both sides #1/2=-cotycscy*(dy)/(dx)# or #(dy)/(dx)=-1/2*1/coty*1/cscy# = #-1/2*1/sqrt(csc^2y-1)*1/cscy# = #-1/2*1/sqrt(x^2/4-1)*2/x# = #-2/(xsqrt(x^2-4))# 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 2293 views around the world You can reuse this answer Creative Commons License