How do you differentiate #f(x) = 1/sqrt(arctan(e^x-1) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer A. S. Adikesavan Sep 25, 2016 #=-e^x/(2((2-2e^x+e^(2x)) arc tan (e^x-1)sqrt(arc tan (e^x-1)))# Explanation: #f'=((arc tan (e^x-1))^(-1/2))'# #=-1/2(arc tan (e^x-1))^(-3/2)(aec tan(e^x-1))'# #=-1/2(arc tan (e^x-1))^(-3/2)(1/(1+(e^x-1)^2))((e^x-1)'# #=-1/2(arc tan (e^x-1))^(-3/2)(1/(1+(e^x-1)^2))(e^x)# #=-e^x/(2((2-2e^x+e^(2x)) arc tan (e^x-1)sqrt(arc tan (e^x-1)))# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1262 views around the world You can reuse this answer Creative Commons License