## Calculus: Early Transcendentals 8th Edition

(a) $$F'(x)=f'(e^x)e^x$$ (b) $$G'(x)=e^{f(x)}f'(x)$$
(a) $$F(x)=f(e^x)$$ The derivative of $F(x)$, according to Chain Rule, would be $$F'(x)=\frac{df(e^x)}{dx}=\frac{df(e^x)}{d(e^x)}\frac{d(e^x)}{dx}$$ We know that $\frac{d(e^x)}{dx}=e^x$ So, $$F'(x)=f'(e^x)e^x$$ (b) $$G(x)=e^{f(x)}$$ The derivative of $G(x)$, according to Chain Rule, would be $$G'(x)=\frac{d[e^{f(x)}]}{dx}=\frac{d[e^{f(x)}]}{df(x)}\frac{df(x)}{dx}$$ Now, $\frac{d[e^{f(x)}]}{df(x)}$ would be just $\frac{d(e^x)}{dx}$. That means $\frac{d[e^{f(x)}]}{df(x)}=e^{f(x)}$ So, $$G'(x)=e^{f(x)}f'(x)$$