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\centerline{MATH 135 -- Calculus 1}
\centerline{The Derivative Product Rule}
\centerline{October 21, 2019}
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\noindent
{\it Background}
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In today's video, we saw that the derivative of a product of functions is given by 
$$(f\cdot g)'(x) = f(x)g'(x) + g(x)f'(x),$$
provided that $f$ and $g$ are each differentiable at $x$.  (In other words, if $f,g$
are differentiable separately, then so is the product, and the derivative of $f\cdot g$
is given by the formula above.)  Today, we want to practice with this new derivative rule
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\noindent
{\it Questions}
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\noindent
\begin{enumerate}
\item[(1)]  Differentiate by the product rule:
\begin{enumerate}
\item[(a)]  $(x^2 + 3x + 1)(x^3 + x + 2)$.  Then, expand out the product, differentiate using 
our more basic rules, and compare your answers.  They should be equal.
\item[(b)]  $(x^2 + x + 1)e^x$.
\item[(c)]  $e^{2x}$  (How can you break this up as an $f(x)g(x)$?)
\item[(d)]  $(x^{1/2} + 3x)(x - 4e^x)$. 
\end{enumerate}
\item[(2)] Differentiate using the product rule ``in two stages:'' 
$$(x^2 + 3x)(e^x + x)(x + x^{-2}).$$
\item[(3)]  Now suppose you have a product $y = f(x)g(x)h(x)$ with {\it three factors}
in general.  If you group the terms like this:
$$f(x)g(x)h(x) = f(x)\cdot (g(x)h(x)),$$
explain why 
$$y' = f(x)(g(x)h(x))' + (g(x)h(x))f'(x).$$
Now work out $(g(x)h(x))'$ by the product rule and rewrite things to see
$$y' = f(x)g(x)h'(x) + f(x)g'(x)h(x) + f'(x)g(x)h(x).$$
\item[(4)]  Continuing from (3), what is the pattern here?  What if you have
a product with any number of factors $y = f_1(x)f_2(x) \cdots f_n(x)$.  What
will the derivative of that product look like?
\end{enumerate}
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