What is the derivative of #ln e^(2x)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Tiago Hands Oct 3, 2016 #y=ln(u)# #:. dy/(du)=1/u=1/(e^(2x))# #u=e^(2x)#, #:. (du)/(dx)=2e^(2x)# Now, using the chain rule... #(dy)/(du)*(du)/(dx)=1/(e^(2x))*2e^(2x)=2# #:. (dy)/(dx)=2# This all worked because: When #y=e^(f(x))# #(dy)/(dx)=f'(x)e^(f(x))# 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 22934 views around the world You can reuse this answer Creative Commons License