Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 81

a.) Prove that if $f$, $g$, and $h$ are differentiable, then $(fgh)' = f'gh+fg'h+fgh'$ by using product rule

$
\begin{equation}
\begin{aligned}
(fgh)' & = [f(gh)]'
&& \text{Group the three functions and assume that we only have two factors}\\
\\
(fgh)' & = f(gh)' + f'(gh)
&& \text{Apply Product rule}\\
\\
(fgh)' & = f(gh'+g'h)+f'gh
&& \text{Apply product rule again in } (gh)'
\end{aligned}
\end{equation}
$


Therefore, $(fgh)' = fgh'+fg'h+f'gh$

b.) Prove that $\displaystyle \frac{d}{dx}[f(x)]^3 = 3[f(x)]^2f'(x)$ by taking $f =g = h$ in part(a)
Let $f^3 = (fgh)$, so we have
$(f^3)' = (fff)'$

Applying Product rule twice we get


$
\begin{equation}
\begin{aligned}
(f^3)' & = f'ff+f(ff)'\\
\\
(f^3)' & = f'ff+f(f'f+f'f)\\
\\
(f^3)' & =f^2f'+f^2f'+f^2f'\\
\\
(f^3)' & = 3f^2f'
\end{aligned}
\end{equation}
$


In other words, $\displaystyle \frac{d}{dx} [f(x)]^3 = 3 [f(x)]^2 f'(x)$

c.) Differentiate $y = (x^4+3x^3+17x+82)^3$ using part(b)
Let $f(x) = x^4 + 3x^3 + 17x + 82$

Using $\displaystyle \frac{d}{dx}[f(x)]^3 = 3[f(x)]^2f'(x)$


$
\begin{equation}
\begin{aligned}
\frac{d}{dx} [f(x)]^3 & = 3(x^4+3x^3+17x+82)^2 \left[ \frac{d}{dx}(x^4)+3\frac{d}{dx}(x^3)+17\frac{d}{dx}(x)+\frac{d}{dx}(82)\right]\\
\\
\frac{d}{dx} [f(x)]^3 & = 3(x^4+3x^3+17x+82)^2 [4x^3+(3)(3x^2)+(17)(1)+0]\\
\\
\frac{d}{dx} [f(x)]^3 & = 3(x^4+3x^3+17x+82)^2 (4x^3+9x^2+17)
\end{aligned}
\end{equation}
$