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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 Quotient Rule{]}}} The derivative of a quotient \[
\frac{f}{g},\]
 is given by\[
\left( \frac{f}{g}\right) '=\frac{f'\cdot g-g'\cdot f}{g^{2}}.\]

\end{thm}
\begin{proof}
We have to do this by the definition, as well. The trick will be a little different.
Watch for it: \begin{eqnarray*}
\left( \frac{f}{g}\right) '(x) & = & \lim _{h\rightarrow 0}\frac{\left( \frac{f}{g}\right) (x+h)-\left( \frac{f}{g}\right) (x)}{h}\\
 & = & \lim _{h\rightarrow 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}\\
 & = & \lim _{h\rightarrow 0}\frac{\left( \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x)g(x+h)}\right) }{h}\\
 & = & \lim _{h\rightarrow 0}\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x)g(x+h)h}\\
 & = & \lim _{h\rightarrow 0}\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x)g(x+h)h}\\
 & = & \lim _{h\rightarrow 0}\left( \frac{f(x+h)g(x)-f(x)g(x)}{g(x)g(x+h)h}+\frac{f(x)g(x)-f(x)g(x+h)}{g(x)g(x+h)h}\right) \\
 & = & \left( \lim _{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\frac{g(x)}{g(x)g(x+h)}-\lim _{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}\frac{f(x)}{g(x)g(x+h)}\right) \\
 & = & \frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^{2}}.
\end{eqnarray*}
\end{proof}

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