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\begin{document}
%MODULE 1: Intro to SAGE


%USER DEFINED COMMANDS
\global\long\def\ds{{\displaystyle }}
 \global\long\def\nsp{\hspace{1em}*{9pt}}
 \global\long\def\lns{\rule{0pt}{11pt} }
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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{Derivatives of logarithmic functions}


\subtitle{In this module we determine the derivative of the logarithmic function
$\ec\ln x$. We also use it to find the derivatives of more difficult
functions.}

\makebeamertitle

\lyxframeend{}\lyxframe{{Defining the problem}}

\begin{question} What are the derivatives of the following functions? 
\begin{itemize}
\item $\ec y=\ln x$ 
\item $\ec y=x^{\sqrt{x}}$ 
\end{itemize}
\end{question}


\lyxframeend{}\lyxframe{{The derivative of $\ln x$}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent We start with the derivative of $\ec\ln x$. Before showing
explicitly what its derivative is, let's take an educated guess using
the SAGElets {}``Tangent Line Guesser'' and {}``Function Factory''.


\pause{}


Using the Tangent Line Guesser, estimate the value of the derivative
of $\ec\ln x$ at the following values of $\ec x$. 


\begin{center}
{\ec%
\begin{tabular}{c|ccccc}
$x$  & $\frac{1}{2}$  & 1  & 2  & 5  & 10 \tabularnewline
\hline 
$(\ln x)'$  &  &  &  &  & \tabularnewline
\end{tabular}}
\par\end{center}


\pause{}
\begin{onlyenv}%{}
<3>

(SAGE may report an error after you change the value of $\ec f$ in
the SAGElet to {\ttcolor ln(x)}. This is expected, because $\ec a$
is initially 0 and $\ec\ln0$ is undefined. Change the value of $\ec a$
and the error should disappear.)

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>

Once you have some values, plug them into the Function Factory.

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<5->

Once you have some values, plug them into the Function Factory. Your
function should look like the one below. You should recognize it;
modify $\ec f$ to match it.
\begin{onlyenv}%{}
<5>

\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/points_for_deriv} 
\end{figure}


\end{onlyenv}%{}
\end{onlyenv}%{}

\pause{}


\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/guess_deriv} 
\end{figure}



\noindent This looks like $\ec1/x$. But is it?

\end{overlayarea}

\lyxframeend{}\lyxframe{{The derivative of $\ln x$}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{onlyenv}%{}
<1>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\ec y & {\ec=\ln x}.
\end{align*}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<2>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\ec e^{y} & {\ec=e^{\ln x}}.
\end{align*}
 \begin{explanation}The exponential function is one-to-one.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\ec e^{y} & {\ec=x}.
\end{align*}
 \begin{explanation}The exponential function and the logarithm are
inverse functions.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\ec e^{y}\cdot y' & {\ec=1}.
\end{align*}
 \begin{explanation}Took the derivative of both sides \emph{with
respect to $x$};\\
 implicit differentation tells us that the derivative of $e^{y}$\\
 with respect to $x$ is $e^{y}\cdot y'$.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<5>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\ec y' & {\ec=e^{-y}}.
\end{align*}
 \begin{explanation}Multiplied both sides by $e^{-y}$.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<6>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\ec y' & {\ec=e^{-\ln x}}.
\end{align*}
 \begin{explanation}Substituted $y=\ln x$---see first line.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<7>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\dims y' & \dims=e^{-\ln x}\\
\ec y' & {\ec=e^{\ln x^{-1}}}.
\end{align*}
 \begin{explanation}Property of logarithms.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<8>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\dims y' & \dims=e^{-\ln x}\\
\dims y' & \dims=e^{\ln x^{-1}}\\
\ec y' & {\ec=x^{-1}}.
\end{align*}
 \begin{explanation}The exponential function and the logarithm are
inverses.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<9>

\noindent To find the derivative of $\ec\ln x$, we will use implicit
differentiation. 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\dims y' & \dims=e^{-\ln x}\\
\dims y' & \dims=e^{\ln x^{-1}}\\
\dims y' & \dims=x^{-1}\\
\ec y' & {\ec=\frac{1}{x}}.
\end{align*}
 \begin{explanation}Property of exponents.\end{explanation}

\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{The derivative of $\ln x$}}

\noindent The preceding argument confirms our initial guess.
\begin{theorem}%{}
The derivative of $\ec\ln x$ is $\ec1/x$.
\end{theorem}%{}

\lyxframeend{}\lyxframe{{Logarithmic differentiation}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent Once we know the derivative of the logarithm, we can use
it to find derivatives of functions where \emph{the variable is in
both the base and the exponent}. We call this technique {\hilite
logarithmic differentiation}.


\pause{}
\begin{onlyenv}%{}
<2>

The second example illustrates the technique. We take the derivative
of 
\begin{align*}
\ec y & {\ec=x^{\sqrt{x}}}.
\end{align*}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>

The second example illustrates the technique. We take the derivative
of 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\ec\ln y & {\ec=\ln x^{\sqrt{x}}}.
\end{align*}
 \begin{explanation}Took the logarithm of both sides.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>

The second example illustrates the technique. We take the derivative
of 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\ln x^{\sqrt{x}}}\\
\ec\ln y & {\ec=\sqrt{x}\ln x}.
\end{align*}
 \begin{explanation}Property of logarithms.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<5>

The second example illustrates the technique. We take the derivative
of 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\ln x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\sqrt{x}\ln x}\\
\ec\frac{1}{y}\cdot y' & {\ec=\frac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\frac{1}{x}}.
\end{align*}
 \begin{explanation}Took the derivative of both sides. By implicit
differentiation and the derivative of the logarithm, $(\ln y)'=\frac{1}{y}\cdot y'$.\\
 The derivative of the right hand side uses the product rule.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<6>

The second example illustrates the technique. We take the derivative
of 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\ln x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\sqrt{x}\ln x}\\
\dims\frac{1}{y}\cdot y' & {\dims=\frac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\frac{1}{x}}\\
\ec y' & {\ec=y\left(\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}\right)}.
\end{align*}
 \begin{explanation}Multiplied $y$ to both sides.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<7>

The second example illustrates the technique. We take the derivative
of 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\sqrt{x}\ln x}\\
\dims\frac{1}{y}\cdot y' & {\dims=\frac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\frac{1}{x}}\\
\dims y' & {\dims=y\left(\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}\right)}\\
\ec y' & {\ec=x^{\sqrt{x}}\left(\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}\right)}.
\end{align*}
 \begin{explanation}Substitued $y=x^{\sqrt{x}}$---see first line.\end{explanation}

\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{Derivatives of other logarithms}}

\noindent For the derivative of a different logarithmic base, we can
reason from the {\hilite base conversion rule} for logarithms that
\[
{\ec\left(\log_{a}x\right)'=\left(\frac{\ln x}{\ln a}\right)'=\frac{1}{\ln a}\left(\ln x\right)'=\frac{1}{\ln a}\frac{1}{x}=\frac{1}{x\ln a}}.
\]
 Thus
\begin{theorem}%{}
\noindent  For any $\ec a\in\mathbb{R}$ where $\ec a>0$ and $\ec a\neq1$,
the derivative of $\ec\log_{a}x$ is $\ec\frac{1}{x\ln a}$. 
\end{theorem}%{}

\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item We determined the derivative of $\ec\ln x$ both geometrically and
precisely, using implicit differentiation. 
\item The derivative $\ec\ln x$ allows us to compute the derivatives of
functions with variables in both the base and the exponent using {\hilite
logarithmic differentiation}. 
\end{itemize}

\lyxframeend{}
\end{document}
