How do you differentiate # g(x) = 2arcsin(e^(3x)) #? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Monzur R. May 26, 2017 #g'(x) = (6e^(3x))/sqrt(1-e^(6x))# Explanation: Let #g(x) = y# #y= 2arcsin(e^(3x))# #1/2y= arcsin(e^(3x))# #sin(1/2y) = e^(3x)# #(1/2cos(1/2y))(dy/dx) = 3e^(3x)# #dy/dx = (6e^(3x))/cos(1/2y)# #cos(1/2y) = sqrt(1-sin^2(1/2y)) = sqrt (1-e^(6x))# #dy/dx = (6e^(3x))/sqrt(1-e^(6x))# 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 1420 views around the world You can reuse this answer Creative Commons License