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\begin{document}
%MODULE: Continuity


%USER DEFINED COMMANDS
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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{The derivative of the exponential function}


\subtitle{In this module we determine the derivative of the logarithmic functions
$\ec e^{x}$ and $\ec a^{x}$, where $\ec a$ is an appropriate constant.}

\makebeamertitle

\lyxframeend{}\lyxframe{{Introduction}}

In this module we determine more efficient ways of computing the derivatives
of \emph{exponential functions}. (Recall that an \emph{exponential
function} has the form $\ec a^{x}$ where the base $\ec a$ is a positive
real number, not equal to 1. An important exponential function uses
the base $\ec e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}\approx2.7183$.)
As with polynomials, we use the precise definition of the derivative.

\noindent %1



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

\begin{question} How do we evaluate the derivative of an exponential
function $\ec f(x)=a^{x}$? \end{question} %2



\lyxframeend{}\lyxframe{{SAGE worksheets}}

\noindent You will need a blank SAGE worksheet for this module.


\lyxframeend{}\lyxframe{{A difficulty with the precise definition}}

\label{pg: defining problem} The precise definition of the derivative
tells us that
\[
{\ec f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}}.
\]
 For an exponential function $\ec f(x)=a^{x}$, this evaluates to
\[
{\ec f'(x)=\lim_{h\rightarrow0}\frac{a^{x+h}-a^{x}}{h}}.
\]
 How do we simplify this expression? We would like a shortcut, so
as to avoid it. %2



\lyxframeend{}\lyxframe{\noindent {The derivative of $e^{x}$}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent We start with a special case, the derivative of $\ec f(x)=e^{x}$.


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

Rewrite the difference quotient found on \RefPage{pg: defining problem}:
\begin{align*}
\ec f'(x) & {\ec=\lim_{h\rightarrow0}\frac{e^{x+h}-e^{x}}{h}}.
\end{align*}


\end{onlyenv}%{}

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

Rewrite the difference quotient found on ~\RefPage{pg: defining
problem}:
\begin{align*}
\dims f'(x) & \dims=\lim_{h\rightarrow0}\frac{e^{x+h}-e^{x}}{h}\\
 & {\ec=\lim_{h\rightarrow0}\frac{e^{x}\cdot e^{h}-e^{x}}{h}}.
\end{align*}
 \begin{explanation}A property of exponents.\end{explanation}

\end{onlyenv}%{}

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

Rewrite the difference quotient found on ~\RefPage{pg: defining
problem}:
\begin{align*}
\dims f'(x) & \dims=\lim_{h\rightarrow0}\frac{e^{x}\cdot e^{h}-e^{x}}{h}\\
 & {\ec=\lim_{h\rightarrow0}\frac{e^{x}(e^{h}-1)}{h}}.
\end{align*}
 \begin{explanation}Factored a common $e^{x}$.\end{explanation}

\end{onlyenv}%{}

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

Rewrite the difference quotient found on ~\RefPage{pg: defining
problem}:
\begin{align*}
\dims f'(x) & \dims=\lim_{h\rightarrow0}\frac{e^{x}(e^{h}-1)}{h}\\
 & {\ec=e^{x}\lim_{h\rightarrow0}\frac{e^{h}-1}{h}}.
\end{align*}
 \begin{explanation}Since $e^{x}$ is constant \emph{with respect
to $h$},\\
 we can factor it from the limit.\end{explanation}

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

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

\noindent So far we have reasoned that if $\ec f(x)=e^{x}$, then
\[
{\ec f'(x)=e^{x}\lim_{h\rightarrow0}\frac{e^{h}-1}{h}}.
\]
 If we could determine the value of the limit, we would be done!


\pause{}


\bigskip{}

\begin{onlyenv}%{}
<2>

We cannot substitute $\ec h=0$ into the expression, because the function
is not continuous at $\ec h=0$ (division by zero). However, you have
learned how to use SAGE to estimate the value of the limit. Use it
now to estimate, \emph{both numerically and graphically}, the value
of the limit.

\end{onlyenv}%{}

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

\emph{Hint 1:} To evaluate it numerically, define the function $\ec f=(e^{h}-1)/h$
and substitute values of $\ec h$ that approach zero. You can use
the commands\\
 \texttt{\ttcolor var('h')}~\\
\texttt{ f = (e\^{}h-1)/h}~\\
\texttt{ $[$round(f(x),4) for x in $[$1,0.5,0.1,0.01,0.001,-0.001,-0.01,-0.1,-0.5,-1$]]$}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4-5>

You should find


\noindent \begin{center}
{\ec%
\begin{tabular}{c|c|c|c|c|c|c}
$h$  & 1  & 0.5  & 0.1  & 0.01  & 0.001  & {\hilite 0} \tabularnewline
\hline 
$f(h)$  & 1.7183  & 1.2974  & 1.0517  & 1.005  & 1.0005  & {\hilite ??} \tabularnewline
\end{tabular}}
\par\end{center}


\noindent \begin{center}
{\ec%
\begin{tabular}{c|c|c|c|c|c|c}
$h$  & {\hilite 0}  & -0.001  & -0.01  & -0.1  & -0.5  & -1 \tabularnewline
\hline 
$f(h)$  & {\hilite ??}  & 0.9995  & 0.995  & 0.9516  & 0.7869  & 0.6321 \tabularnewline
\end{tabular}}
\par\end{center}

\end{onlyenv}%{}

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

\noindent It \emph{looks} as if $\ec f(h)\rightarrow1$ when $\ec h\rightarrow0$.
See if a graph gives the same intuition.

\end{onlyenv}%{}

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

You should see a graph that looks like this: 
\begin{figure}[h!]
 

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



\noindent The graph appears to confirm our belief that the limit is
1.

\end{onlyenv}%{}

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

We will assume that our intuition is correct. Thus
\[
\ec f'(x)=e^{x}\cdot1=e^{x}.
\]
 In other words, $\ec e^{x}$ \emph{is its own derivative}!

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

\lyxframeend{}\lyxframe{\noindent {BIG-TIME FACT}}
\begin{theorem}%{}
[($\ec e^{x}$ is its own derivative)] If $\ec f(x)=e^{x}$, then
$\ec f'(x)=e^{x}$. Written another way,
\[
{\ec y=e^{x}\qquad\Longrightarrow\qquad y'=y}.
\]
 
\end{theorem}%{}
\begin{explanation}We have argued intuitively that 
\[
\lim_{h\rightarrow0}\frac{e^{h}-1}{h}=1,
\]
but have not \emph{proved} it. It \emph{can} be proved, using the
definition
\[
{e^{h}=\lim_{x\rightarrow\infty}\left(1+\frac{h}{x}\right)^{x}},
\]
 using an argument similar to (but harder than) the one for the derivative
of a monomial. \end{explanation}


\lyxframeend{}\lyxframe{{Example}}
\begin{overlayarea}
{\textwidth}{6in}

\begin{question}What is the derivative of $\ec e^{4x-3}$?\end{question} 


\pause{}


\noindent We can use the \emph{\hilite Chain Rule} along with the
derivative of $\ec e^{x}$ to answer this. Remember that the Chain
Rule works with a composition of functions, where there is an {}``inside''
and an {}``outside'' function.


\pause{}


\noindent Here the {}``inside'' function is $\ec u(x)=4x-3$, and
the {}``outside'' function is $\ec g(u)=e^{u}$.


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

From the Chain Rule,
\begin{align*}
\ec\frac{d}{dx}e^{4x-3} & {\ec=\frac{dg}{du}\cdot\frac{du}{dx}}.
\end{align*}


\end{onlyenv}%{}

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

From the Chain Rule,
\begin{align*}
\dims\frac{d}{dx}e^{4x-3} & {\dims=\frac{dg}{du}\cdot\frac{du}{dx}}\\
 & {\ec=e^{u}\cdot4}.
\end{align*}
 \begin{explanation}Substituted for the derivatives.\end{explanation}

\end{onlyenv}%{}

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

From the Chain Rule,
\begin{align*}
\dims\frac{d}{dx}e^{4x-3} & {\dims=e^{u}\cdot4}\\
 & {\ec=4e^{4x-3}}.
\end{align*}
 \begin{explanation}Substituted for $u$.\end{explanation}

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

\lyxframeend{}\lyxframe{\noindent {The derivative of $a^{x}$, for arbitrary $a$}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent To determine the derivative of $\ec a^{x}$ for other bases,
we use an algebraic technique along with \emph{\hilite the Chain
Rule}.


\bigskip{}



\pause{}


First the algebraic technique. Recall that $\ec e^{x}$ and $\ec\ln(x)$
are inverse functions. That is, $\ec x=e^{\ln x}$ for \emph{any}
$\ec x$.


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

In that case, we can say that 
\begin{align*}
\ec a^{x} & \ec=e^{\ln\left(a^{x}\right)}
\end{align*}
 \begin{explanation}Inverse functions.\end{explanation}

\end{onlyenv}%{}

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

In that case, we can say that 
\begin{align*}
\dims a^{x} & \dims=e^{\ln\left(a^{x}\right)}\\
 & \ec=e^{x\ln a}
\end{align*}
 \begin{explanation}Property of logarithms.\end{explanation}

\end{onlyenv}%{}

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

In that case, we can say that 
\begin{align*}
\dims a^{x} & \dims=e^{x\ln a}\\
 & \ec=\left(e^{x}\right)^{\ln a}
\end{align*}
 \begin{explanation}Property of exponents.\end{explanation}

\end{onlyenv}%{}

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

From the algebraic technique, we have $\ec a^{x}=\left(e^{x}\right)^{\ln a}$.
Now we'll apply the Chain Rule. Here the {}``inside'' is $\ec u(x)=e^{x}$
and the {}``outside'' is $\ec g(u)=u^{\ln a}$.

\end{onlyenv}%{}

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

Thus 
\begin{align*}
\ec\frac{d}{dx}\left(e^{x}\right)^{\ln a} & {\ec=\ln a\cdot\left(e^{x}\right)^{\ln a-1}\cdot\left(e^{x}\right)}.
\end{align*}
 \begin{explanation}The Chain Rule, and substitution.\end{explanation}

\end{onlyenv}%{}

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

Thus 
\begin{align*}
\dims\frac{d}{dx}\left(e^{x}\right)^{\ln a} & \dims=\ln a\cdot\left(e^{x}\right)^{\ln a-1}\cdot\left(e^{x}\right)\\
 & {\ec=\ln a\cdot\left(e^{x}\right)^{\ln a\cancelto{0}{-1+1}}}.
\end{align*}
 \begin{explanation}A property of exponents \emph{that have the same
base}.\\
 In this case the base is $\ec e^{x}$.\end{explanation}

\end{onlyenv}%{}

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

Thus 
\begin{align*}
\dims\frac{d}{dx}\left(e^{x}\right)^{\ln a} & \dims=\ln a\cdot\left(e^{x}\right)^{\ln a\cancelto{0}{-1+1}}\\
 & \ec=\ln a\cdot e^{x\ln a}
\end{align*}
 \begin{explanation}A property of exponents.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<10>

Thus 
\begin{align*}
\dims\frac{d}{dx}\left(e^{x}\right)^{\ln a} & \dims=\ln a\cdot e^{x\ln a}\\
 & \ec=\ln a\cdot e^{\ln a^{x}}
\end{align*}
 \begin{explanation}A property of logarithms.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<11>

Thus 
\begin{align*}
\dims\frac{d}{dx}\left(e^{x}\right)^{\ln a} & \dims=\ln a\cdot e^{\ln a^{x}}\\
 & {\ec=\ln a\cdot a^{x}}.
\end{align*}
 \begin{explanation}$e^{x}$ and $\ln x$ are inverse functions.\end{explanation}

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

\lyxframeend{}\lyxframe{{The derivatives of exponential functions}}

\noindent We have finally found the derivatives we wanted!
\begin{theorem}%{}
\noindent  
\begin{itemize}
\item The derivative of $\ec e^{x}$ is itself. 
\item The derivative of $\ec a^{x}$ is $\ec a^{x}\ln a$. 
\end{itemize}
\end{theorem}%{}
You should commit these to memory.


\lyxframeend{}\lyxframe{{Examples}}

\noindent As with other derivative {}``shortcuts'', we can use them
with the properties of the derivative. For example,
\begin{align*}
\ec\frac{d}{dx}\left(2e^{x}\right) & \ec=2\frac{d}{dx}e^{x}=2x\\
\ec\frac{d}{dx}\left(xe^{x}\right) & \ec=e^{x}+xe^{x}\\
\ec\frac{d}{dx}\left(\frac{2^{x}}{x^{2}}\right) & \ec=\frac{2^{x}(\ln2)x^{2}-2^{x+1}x}{x^{4}}\\
\ec\frac{d}{dx}\left(e^{x^{2}}\right) & \ec=2x\cdot e^{x^{2}}
\end{align*}
 These illustrate the constant multiple rule, the product rule, the
quotient rule, and the Chain Rule, respectively.


\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item We found a shortcut for the derivative of $\ec e^{x}$. Interestingly,
$\ec e^{x}$ is its own derivative.\setlength{\shadeboxwidth}{4.5in}

\begin{theorem}%{}
 If $\ec f(x)=e^{x}$, then $\ec f'(x)=e^{x}$. Written another way,
\[
{\ec y=e^{x}\qquad\Longrightarrow\qquad y'=y}.
\]
 
\end{theorem}%{}
\item To find this shortcut, we used the precise definition of the derivative
and an intuitive estimate of the limit
\[
{\ec\lim_{h\rightarrow0}\frac{e^{h}-1}{h}=1}.
\]
 
\item We used the shortcut to find a shortcut for the derivative of $\ec a^{x}$
where the base $\ec a$ is any positive real number, not equal to
one.

\begin{theorem}%{}
The derivative of $\ec a^{x}$ is $\ec a^{x}\ln a$. 
\end{theorem}%{}
\item To find this second shortcut, we made use of properties of exponents
and logarithms, along with the Chain Rule. 
\item These shortcuts can be combined with the properties of derivatives. 
\end{itemize}

\lyxframeend{}
\end{document}
