How do you differentiate #y=csc^-1(e^x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer VinÃcius Ferraz Jul 9, 2017 #dy/dx = - tan (csc^-1 (e^x))# Explanation: #csc y = e^x => ln (1/sin y) = x# #dx/dy = sin y * (-1/sin^2 y) * cos y# #dy/dx = (dx/dy)^-1 = -sin y / cos y = - tan (csc^-1 (e^x))# 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 3573 views around the world You can reuse this answer Creative Commons License