How do I find the derivative of #y= ln(1 + e^(2x)) #?

1 Answer
Jan 1, 2016

Using chain rule, which states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

Explanation:

We'll apply that logic for both the #ln# and the #e^(2x)#, which demand chain rule to become differentiable.

Renaming #u=1+e^(2x)# we can differentiate the #ln# and by renaming #v=2x# we can differentiate the #e^(2x)#.

#(dy)/(dx)=1/u(2e^(2x))=(2e^(2x))/(1+e^(2x))#