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\newtheorem{theorem}{Theorem}
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\newtheorem{conclusion}[theorem]{Conclusion}
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\lhead{\color{blue} \Large Lecture Notes}
\chead{\color{black} \LARGE The Quotient Rule}
\rhead{\large page   \ \thepage}
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\lfoot{\small   \copyright $\;$ copyright  Hidegkuti 2013}
\rfoot{\small   Last revised: October 22, 2013}
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\qquad Theorem: \ Suppose that $f$ and $g$ are differentiable functions with 
$g\not=0$. \ Then 
\begin{equation*}
\left( \dfrac{f}{g}\right) ^{\prime }=\dfrac{f^{\prime }g-fg^{\prime }}{g^{2}%
}
\end{equation*}%
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\begin{eqnarray*}
\left( \dfrac{f}{g}\right) ^{\prime } &=&\lim\limits_{h\rightarrow 0}\dfrac{%
\dfrac{f}{g}\left( x+h\right) -\dfrac{f}{g}\left( x\right) }{h}%
=\lim\limits_{h\rightarrow 0}\dfrac{\dfrac{f\left( x+h\right) }{g\left(
x+h\right) }-\dfrac{f\left( x\right) }{g\left( x\right) }}{h}%
=\lim\limits_{h\rightarrow 0}\dfrac{~~\dfrac{f\left( x+h\right) g\left(
x\right) -f\left( x\right) g\left( x+h\right) }{g\left( x\right) g\left(
x+h\right) }~~}{h} \\
&=&\lim\limits_{h\rightarrow 0}\dfrac{1}{h}\cdot \dfrac{f\left( x+h\right)
g\left( x\right) -f\left( x\right) g\left( x+h\right) }{g\left( x\right)
g\left( x+h\right) }=\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right)
g\left( x\right) -f\left( x\right) g\left( x+h\right) }{hg\left( x\right)
g\left( x+h\right) }
\end{eqnarray*}

We smuggle in $f\left( x\right) g\left( x\right) $%
\begin{eqnarray*}
\left( \dfrac{f}{g}\right) ^{\prime } &=&\lim\limits_{h\rightarrow 0}\dfrac{%
f\left( x+h\right) g\left( x\right) -f\left( x\right) g\left( x\right)
+f\left( x\right) g\left( x\right) -f\left( x\right) g\left( x+h\right) }{%
hg\left( x\right) g\left( x+h\right) } \\
&=&\lim\limits_{h\rightarrow 0}\left( \dfrac{f\left( x+h\right) g\left(
x\right) -f\left( x\right) g\left( x\right) }{hg\left( x\right) g\left(
x+h\right) }+\dfrac{f\left( x\right) g\left( x\right) -f\left( x\right)
g\left( x+h\right) }{hg\left( x\right) g\left( x+h\right) }\right)
\end{eqnarray*}%
We factor out $g\left( x\right) $ in the first term and $f\left( x\right) $
in the second. 
\begin{eqnarray*}
\left( \dfrac{f}{g}\right) ^{\prime } &=&\lim\limits_{h\rightarrow 0}\left( 
\dfrac{g\left( x\right) \left[ f\left( x+h\right) -f\left( x\right) \right] 
}{hg\left( x\right) g\left( x+h\right) }+\dfrac{f\left( x\right) \left[
g\left( x\right) -g\left( x+h\right) \right] }{hg\left( x\right) g\left(
x+h\right) }\right) \\
&=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{%
hg\left( x+h\right) }+\lim\limits_{h\rightarrow 0}\dfrac{f\left( x\right) %
\left[ g\left( x\right) -g\left( x+h\right) \right] }{hg\left( x\right)
g\left( x+h\right) } \\
&=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{%
h}\cdot \lim\limits_{h\rightarrow 0}\dfrac{1}{g\left( x+h\right) }+f\left(
x\right) \lim\limits_{h\rightarrow 0}\dfrac{-\left[ g\left( x+h\right)
-g\left( x\right) \right] }{hg\left( x\right) g\left( x+h\right) } \\
&=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{%
h}\cdot \lim\limits_{h\rightarrow 0}\dfrac{1}{g\left( x+h\right) }+f\left(
x\right) \lim\limits_{h\rightarrow 0}\left( -\dfrac{g\left( x+h\right)
-g\left( x\right) }{h}\right) \lim\limits_{h\rightarrow 0}\dfrac{1}{g\left(
x\right) g\left( x+h\right) } \\
&=&~~~~~~~~~f^{\prime }\left( x\right) ~~~~~~~~~~\cdot ~~~~~~~~\dfrac{1}{%
g\left( x\right) }~~~~~+f\left( x\right) \cdot ~~\left( -g^{\prime }\left(
x\right) \right) ~~~~\cdot ~~~~~~~\dfrac{1}{\left( g\left( x\right) \right)
^{2}}
\end{eqnarray*}%
This last conclusion is correct because of the following facts:\medskip

$\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}%
=f^{\prime }\left( x\right) $ by the definition of the derivative\medskip

$\lim\limits_{h\rightarrow 0}\dfrac{1}{g\left( x+h\right) }=\dfrac{1}{%
\lim\limits_{h\rightarrow 0}g\left( x+h\right) }=\dfrac{1}{g\left( x\right) }
$ because $g$ is continuous at $x$\medskip

$\lim\limits_{h\rightarrow 0}\left( -\dfrac{g\left( x+h\right) -g\left(
x\right) }{h}\right) =-\lim\limits_{h\rightarrow 0}\dfrac{g\left( x+h\right)
-g\left( x\right) }{h}=-g^{\prime }\left( x\right) $ by properties of limits
and the definition of the derivative.\medskip

$\lim\limits_{h\rightarrow 0}\dfrac{1}{g\left( x\right) g\left( x+h\right) }=%
\dfrac{1}{\lim\limits_{h\rightarrow 0}g\left( x\right) g\left( x+h\right) }=%
\dfrac{1}{g\left( x\right) \lim\limits_{h\rightarrow 0}g\left( x+h\right) }=%
\dfrac{1}{\left( g\left( x\right) \right) ^{2}}$ because $g$ is
continuous.\bigskip

Thus we have that

\begin{equation*}
\left( \dfrac{f}{g}\right) ^{\prime }\left( x\right) =\dfrac{f^{\prime
}\left( x\right) }{g\left( x\right) }+\dfrac{f\left( x\right) \left(
-g^{\prime }\left( x\right) \right) }{\left( g\left( x\right) \right) ^{2}}=%
\dfrac{f^{\prime }\left( x\right) g\left( x\right) }{\left( g\left( x\right)
\right) ^{2}}+\dfrac{f\left( x\right) \left( -g^{\prime }\left( x\right)
\right) }{\left( g\left( x\right) \right) ^{2}}=\dfrac{f^{\prime }\left(
x\right) g\left( x\right) -f\left( x\right) g^{\prime }\left( x\right) }{%
\left( g\left( x\right) \right) ^{2}}
\end{equation*}%
\medskip \medskip \medskip

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