## Calculus: Early Transcendentals 8th Edition

(a) First, set aside $(fg)$ and apply the Product Rule for $(fgh)'$ with 2 elements $(fg)$ and $h$. Then apply the Product Rule one more time for $(fg)'$ with 2 elements $f$ and $g$. (b) First, we find that $f^3=fgh$ if $f=g=h$. Then we can apply the result from part a). Remember that in part b), $f=g=h$ (c) $$\frac{dy}{dx}=3e^{3x}$$
(a) $f$, $g$ and $h$ are differentiable, then $f'$, $g'$ and $h'$ exist. Consider $(fgh)'$ $$(fgh)'=[(fg)\times h]'$$ Apply the Product Rule, $$(fgh)'=(fg)'h+(fg)h'$$ Apply the Product Rule again for $(fg)'$, $$(fgh)'=(f'g+fg')h+fgh'$$ $$(fgh)'=f'gh+fg'h+fgh'$$ The statement has been proved. (b) Taking $f=g=h$, we can see that $$fgh=f^3$$ So, $$(f^3)'=(fgh)'=f'gh+fg'h+fgh'$$ However, since $f=g=h$, $$(f^3)'=f'ff+ff'f+fff'$$ $$(f^3)'=3f'ff=3f'f^2$$ Therefore, $$\frac{d}{dx}[f(x)]^3=3[f(x)]^2f'(x)$$ The statement, as a result, has been proved. (c) $$y=e^{3x}=(e^x)^3$$ So, $$\frac{dy}{dx}=\frac{d}{dx}(e^x)^3$$ $$\frac{dy}{dx}=3(e^x)^2(e^x)'$$ $$\frac{dy}{dx}=3e^{2x}e^x=3e^{3x}$$