## Calculus: Early Transcendentals 8th Edition

$f'(x) = g'(e^x)\cdot e^x$
Let $h(x) = e^x$ $h'(x) = e^x$ We can find $f'$: $f(x) = g(e^x)$ $f(x) = g(h(x))$ $f'(x) = g'(h(x))~\cdot h'(x)$ $f'(x) = g'(e^x)\cdot e^x$