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


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