calculus
posted by AHKSIJ .
Calculate f" for f(x)=g(e^(2x)), where g is a function defined for all real numbers & g admits second order derivative.

f = g(e^2x)
f' = 2e^2x g'
f'' = 4e^2xg' + 2e^2xg''
example:
g(x) = x sinx
f(x) = g(e^2x) = e^2x sin(e^2x)
g' = sinx + xcosx
g'' = cosx + cosx  xsinx
= 2cosx  xsinx
f' = 2e^2x sin(e^2x) + e^2xcos(e^2x)*2e^2x
= 2e^2x(sin(e^2x) + e^2x cos(e^2x))
Letting g(e^2x) = g(u)
f' = 2e^2x g'(u)
I have expanded out
4e^2xg'(u) + 2e^2xg''(u)
and it checks out. Have lots of blank paper!