Mark M. answered 03/12/21
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
g(x) = ef(x)
g'(x) = ef(x) f'(x) So, g'(1) = ef(1) f'(1) = 3e2
--------------------------------
h(x) = f(ex)
h'(x) = f'(ex) (ex)' = f(ex) (ex) So, h'(0) = f'(e0) (e0) = f'(1) (1) = 3
Mark M.
03/13/21
Conor E.
how did you get the double (e^x)(e^x) in h'(x)03/12/21