How do you differentiate #y=csc^-1(x/2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Alan N. Nov 30, 2016 #dy/dx=(-2)/(x^2sqrt(1-4/x^2)# Explanation: #y=csc^-1(x/2)# #cscy = x/2# #siny = 2/x# #cosy dy/dx = -2/x^2# (Implicit differentiation and Power rule) #dy/dx = -2/x^2 * 1/cosy# Since #cos^2y + sin^2y =1# #cos^2y = 1-sin^2y = 1-(2/x)^2# #cosy = sqrt(1-4/x^2)# #dy/dx = (-2)/(x^2sqrt(1-4/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 3407 views around the world You can reuse this answer Creative Commons License