## Calculus: Early Transcendentals 8th Edition

$$r'(1)=120$$
$$r(x)=f(g(h(x)))$$ The derivative of $r(x)$ is $$r'(x)=\frac{df(g(h(x)))}{dx}$$ Apply the Chain Rule, we have $$r'(x)=\frac{df(g(h(x)))}{dg(h(x))}\frac{dg(h(x))}{dh(x)}\frac{dh(x)}{dx}$$ $$r'(x)=f'(g(h(x)))g'(h(x))h'(x)$$ Therefore, $$r'(1)=f'(g(h(1)))g'(h(1))h'(1)$$ First, $h(1)=2$ and $h'(1)=4$ $$r'(1)=f'(g(2))g'(2)\times4$$ Next, $g(2)=3$ and $g'(2)=5$ $$r'(1)=f'(3)\times5\times4=20f'(3)$$ Finally, $f'(3)=6$ $$r'(1)=20\times6=120$$