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\makeatletter


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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Differentiation Rules}

\begin{thm}
\textbf{\textup{{[}The Product Rule{]}}} The derivative of a product \( fg \),
is given by \[
(f\cdot g)'=f'\cdot g+f\cdot g'.\]

\end{thm}
\begin{proof}
We have to do this by the definition. As we always start off, \[
(f\cdot g)'(x)=\lim _{h\rightarrow 0}\frac{(f\cdot g)(x+h)-(f\cdot g)(x)}{h}=\lim _{h\rightarrow 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}.\]

\end{proof}
Now comes the trick. We need to break the difference quotient up into pieces,
to see the derivatives of \( f \) and \( g \) separately. The trick is to
subtract and add \( f(x)g(x+h) \), so that: \begin{eqnarray*}
(f\cdot g)'(x) & = & \lim _{h\rightarrow 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\
 & = & \lim _{h\rightarrow 0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}.
\end{eqnarray*}
Now we break this mess up into two different fractions,\begin{eqnarray*}
(f\cdot g)'(x) & = & \cdots \\
 & = & \lim _{h\rightarrow 0}\frac{f(x+h)g(x+h)-f(x)g(x+h)}{h}+\lim _{h\rightarrow 0}\frac{f(x)g(x+h)-f(x)g(x)}{h}.
\end{eqnarray*}
and we factor out, in the first limit, the coommon factor of \( g(x+h) \) and
in the second limit we factor out the factor of \( f(x) \). After that, the
derivatives of \( f \) and \( g \) appear, and we can take those limits since
we assume we know what the derivatives of \( f \) and \( g \) are:\begin{eqnarray*}
(f\cdot g)'(x) & = & \cdots \\
 & = & \lim _{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}g(x+h)+\lim _{h\rightarrow 0}f(x)\frac{g(x+h)-g(x)}{h}\\
 & = & \lim _{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\lim _{h\rightarrow 0}g(x+h)+f(x)\lim _{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}\\
 & = & f'(x)g(x)+f(x)g'(x).
\end{eqnarray*}

\end{document}
