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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{Derivatives of other Exponential and Logarithmic functions}


\subsubsection{Derivatives of exponential functions}

\begin{example}
If \[
f(x)=\frac{(x^{2}+3)^{4}e^{3x}}{(x-3)(2x+5)^{2}},\]
 find \( f'(x) \).
\end{example}

\subsubsection*{Solution}

\begin{eqnarray*}
\ln (f(x)) & = & \ln \left( \frac{(x^{2}+3)^{4}e^{3x}}{(x-3)(2x+5)^{2}}\right) \\
 & = & \ln \left( (x^{2}+3)^{4}e^{3x}\right) -\ln \left( (x-3)(2x+5)^{2}\right) \\
 & = & 4\ln \left( x^{2}+3\right) +3x\ln \left( e\right) -\ln \left( x-3\right) -2\ln \left( 2x+5\right) \\
 & = & 4\ln \left( x^{2}+3\right) +3x-\ln \left( x-3\right) -2\ln \left( 2x+5\right) ,
\end{eqnarray*}
so\begin{eqnarray*}
\frac{d}{dx}\ln (f(x)) & = & \left( 4\ln \left( x^{2}+3\right) +3x-\ln \left( x-3\right) -2\ln \left( 2x+5\right) \right) '\\
 & = & \frac{8x}{x^{2}+3}+3-\frac{1}{x-3}-\frac{4}{2x+5},
\end{eqnarray*}
which means that \begin{eqnarray*}
\frac{df}{dx} & = & f(x)\left( \frac{d}{dx}\ln (f(x))\right) \\
 & = & \left( \frac{(x^{2}+3)^{4}e^{3x}}{(x-3)(2x+5)^{2}}\right) \left( \frac{8x}{x^{2}+3}+3-\frac{1}{x-3}-\frac{4}{2x+5}\right) .
\end{eqnarray*}
This may not be terribly pretty, but it is simpler than doing it directly.
\end{document}
