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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}

Any other exponential function, that is, any function \( a^{x} \) where \( a \)
is not \( e \) -- this shows my prejudice in favor of \( e \) when dealing
with exponentials -- can be interpreted in terms of \( e^{x} \), and so its
derivatives can as well.

If \( f(x)=a^{x} \), then, since \[
a=e^{\ln (a)},\]
\begin{eqnarray*}
f(x) & = & a^{x}\\
 & = & \left( e^{\ln (a)}\right) ^{x}\\
 & = & e^{\ln (a)\, x}\\
 & = & e^{x\, \ln (a)}.
\end{eqnarray*}
No derivatives, yet. But, to compute the derivative of \( f(x) \), you use
the chain rule and the last version above for \( f(x) \):\begin{eqnarray*}
f'(x) & = & \left( a^{x}\right) '\\
 & = & \left( e^{x\, \ln (a)}\right) '\\
 & = & e^{x\, \ln (a)}(x\, \ln (a))'\\
 & = & \ln (a)\, a^{x}.
\end{eqnarray*}
So, this is no big deal. But it does show why \( e \) is so useful for calculus;
had we stuck with our biological preference for base 10, we'd have to remember
that \[
\left( 10^{x}\right) '=2.3026\, 10^{x}.\]



\subsubsection{Logarithms.}

For other bases, the formula is simply altered by the rule \( \log _{b}(x)=\frac{\ln x}{\ln b} \),
\[
\left( \log _{b}(x)\right) '=\frac{1}{x\, \ln b}.\]
This is analogous to the formula for \( \left( b^{x}\right) '=b^{x}\ln b \),
which we just derived. 

\begin{example}
\begin{eqnarray*}
\frac{d}{dx}\ln \left( \frac{x+1}{\sqrt{x-2}}\right)  & = & \frac{d}{dx}\left( \ln \left( x+1\right) -\ln \left( \sqrt{x-2}\right) \right) \\
 & = & \frac{d}{dx}\left( \ln \left( x+1\right) -\frac{1}{2}\ln \left( x-2\right) \right) \\
 & = & \frac{1}{x+1}-\frac{1}{2(x-2)}.
\end{eqnarray*}

\end{example}
Actually, the technique used here, using the rules of logarithms to simplify
the expression before the derivative is computed, is sometimes useful. It is
referred to as logarithmic differentiation, and has the advantage of replacing
hard derivative rules (like the quotient rule) with easier ones, like the sum
rule. 

The general idea is that, if \( y=f(x) \) is a {}``messy{}'' expression,
then perhaps \( \ln (f(x)) \) can be simplified, then differentiated, \[
\frac{d}{dx}\ln (f(x))=\frac{1}{f(x)}\frac{df}{dx}.\]
But, the assumption is that the left-hand side of this is easer to compute than
the right, so you find \( f'(x) \) by computing the left-hand side, and:\[
f(x)\frac{d}{dx}\ln (f(x))=\frac{df}{dx}.\]


\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{xca}
Find \[
\left( \frac{\sqrt{x^{2}+3x+1}\left( x^{2}+3\right) ^{2}}{(3x+2)^{4}}\right) '\]
 using implicit differentiation:
\end{xca}
\textbf{\emph{Answer}}

\begin{example}
\begin{eqnarray*}
\left( x^{\sin x}\right) ' & = & \left( \left( e^{\ln x}\right) ^{\sin x}\right) '\\
 & = & \left( e^{\ln x\, \sin x}\right) '\\
 & = & e^{\ln x\, \sin x}(\frac{\sin x}{x}+\ln x\, \cos x)\\
 & = & x^{\sin x}(\frac{\sin x}{x}+\ln x\, \cos x).
\end{eqnarray*}

\end{example}
\begin{xca}
Find \[
\left( x^{x^{x}}\right) '=\]
\end{xca}

\end{document}
