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\begin{document}
%MODULE: Continuity


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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{Properties of derivatives}


\subtitle{As with limits, we do not want to spend a lot of time with the precise
definition of the derivative, nor with graphical or numerical approximations.
We prefer to develop properties (or {}``shortcuts'') that allow
us to skip the tedious algebra and jump directly into problem solving. }

\makebeamertitle
%1



\lyxframeend{}\lyxframe{Defining the problem}

\noindent We would like properties of the derivative that are similar
to properties of limits: 
\begin{itemize}
\item the derivative of a constant; 
\item the derivative of a constant multiple; 
\item the derivative of a sum or a difference of functions; 
\item the derivative of a product of functions; and 
\item the derivative of a quotient of functions. 
\end{itemize}
%2



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

\noindent You will not need a SAGE worksheet for this module.


\lyxframeend{}\lyxframe{{The derivative \emph{as a function}}}

Until now we have mostly considered the derivative \emph{at a point}.
Recall the definition:
\begin{definition}%{}
[(The derivative at a point)] The derivative of a function $\ec f$
at $\ec x=a$ is 
\begin{itemize}
\item $\ec{\displaystyle f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}}$,
or equivalently 
\item $\ec{\displaystyle f'(a)=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}}$. 
\end{itemize}
\end{definition}%{}

\pause{}

\noindent If $\ec f$ is differentiable at every point in its domain,
then we can consider the derivative \emph{as a function}; that is,
a formula that describes $\ec f'(x)$ where $\ec x$ is a \emph{variable}
instead of a \emph{constant}. Take the second definition of the derivative
at a point, and let the point $\ec a$ vary. What does the definition
become?


\lyxframeend{}\lyxframe{The derivative \emph{as a function}}
\begin{definition}%{}
[(The derivative at a point)]The derivative of a function $f$ at
$x=a$ is\ldots{} 
\begin{itemize}
\item ${\displaystyle f'(a)=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}}$. 
\end{itemize}
\end{definition}%{}
\noindent Take the second definition of the derivative at a point,
and let the point $\ec a$ vary. What does the definition become?
\begin{definition}%{}
[(The derivative as a function)] If a function $\ec f$ is differentiable
at every point in its domain, then \emph{\hilite the derivative of
$\ec f$} is
\[
{\ec f'(x)=\lim_{x\rightarrow a}\frac{f(x+h)-f(x)}{h}}.
\]
 
\end{definition}%{}
We simply replaced $\ec a$, which is usually used to represent a
constant, with $\ec x$, which is usually used to represent a variable.


\lyxframeend{}\lyxframe{{The derivative of a constant}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent Suppose that $\ec f(x)=c$ where $\ec c\in\mathbb{R}$ is
a constant; that is, $\ec f(x)$ has the same value regardless of
the value of $\ec x$. What is the derivative?


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

We can make a guess as to what the derivative \emph{should} be by
looking at the graph of a constant, for example $\ec f(x)=2$: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.2]{figures/cplot} 
\end{figure}



\noindent The graph is a horizontal line. Any line tangent to this
graph should also be horizontal. The slope of a horizontal line is
zero, so we can venture that
\[
{\ec f'(x)=0}.
\]


\end{onlyenv}%{}

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

The precise definition of the derivative verifies this for us.
\begin{align*}
\ec f'(x) & {\ec=\lim_{x\rightarrow0}\frac{f(x+h)-f(x)}{h}}.
\end{align*}


\end{onlyenv}%{}

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

The precise definition of the derivative verifies this for us.
\begin{align*}
\dims f'(x) & \dims=\lim_{x\rightarrow0}\frac{f(x+h)-f(x)}{h}\\
 & {\ec=\lim_{x\rightarrow0}\frac{c-c}{h}}.
\end{align*}
 \begin{explanation}$f(x)=c$ \emph{regardless of} the value of $x$.\end{explanation}

\end{onlyenv}%{}

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

The precise definition of the derivative verifies this for us.
\begin{align*}
\dims f'(x) & \dims=\lim_{x\rightarrow0}\frac{f(x+h)-f(x)}{h}\\
 & {\ec=\lim_{x\rightarrow0}\frac{\cancelto{0}{c-c}}{h}}.
\end{align*}


\end{onlyenv}%{}

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

The precise definition of the derivative verifies this for us.
\begin{align*}
\dims f'(x) & \dims=\lim_{x\rightarrow0}\frac{\cancelto{0}{c-c}}{h}\\
 & \ec=\lim_{x\rightarrow0}0\\
 & {\ec=0}.
\end{align*}
 \begin{explanation}The limit of a constant---here, 0---is the constant.\end{explanation}

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

\lyxframeend{}\lyxframe{{Properties of derivatives}}

\noindent We have our first property:
\begin{theorem}%{}
[(Properties of derivatives)] Let $\ec f$ be a function. 
\begin{itemize}
\item If $\ec f(x)$ is constant, then $\ec f'(x)=0$. 
\end{itemize}
\end{theorem}%{}

\lyxframeend{}\lyxframe{{The derivative of a constant multiple}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent We move next to the derivative of a constant multiple. Rather
than make an intuitive argument based on \emph{geometry} here, we
can make an intuitive argument based on \emph{properties of limits}.
Let $\ec f$ be a function, $\ec c\in\mathbb{R}$ a constant, and
$\ec g(x)=cf(x)$. Here $\ec g$ is a \emph{constant multiple} of
$\ec f$.


\pause{}


\bigskip{}

\begin{onlyenv}%{}
<2>

In an earlier module, we considered the derivatives of $\ec\sin x$
and $\hilite4\ec\sin x$ at $\ec x=1$. We found that the first derivative
was $\ec0.54$, and the second was $\ec2.16=\hilite4\ec\cdot0.54$.
This suggests that
\[
{\ec\left(g\right)'=cf'(x)}.
\]
(Here $\ec f(x)=\sin x$ and $\ec c=4$.)

\end{onlyenv}%{}

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

By definition, a derivative is a kind of limit, and from the limit
of a constant multiple we expect a constant multiple of the limit:
\[
{\ec\left(g\right)'=cf'(x)}.
\]


\end{onlyenv}%{}

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

\noindent A close examination of the matter using the precise definition
confirms this: 
\begin{align*}
\ec g'(x) & {\ec=\lim_{h\rightarrow0}\frac{g(x+h)-g(x)}{h}}.
\end{align*}


\end{onlyenv}%{}

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

\noindent A close examination of the matter using the precise definition
confirms this. 
\begin{align*}
\dims g'(x) & \dims=\lim_{h\rightarrow0}\frac{g(x+h)-g(x)}{h}\\
 & {\ec=\lim_{h\rightarrow0}\frac{cf(x+h)-cf(x)}{h}}.
\end{align*}
 \begin{explanation}Substituted $g(x)=cf(x)$.\end{explanation}

\end{onlyenv}%{}

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

\noindent A close examination of the matter using the precise definition
confirms this. 
\begin{align*}
\dims g'(x) & \dims=\lim_{h\rightarrow0}\frac{cf(x+h)-cf(x)}{h}\\
 & {\ec=\lim_{h\rightarrow0}\frac{c\left[f(x+h)-f(x)\right]}{h}}.
\end{align*}
 \begin{explanation}Factored a common $c$.\end{explanation}

\end{onlyenv}%{}

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

\noindent A close examination of the matter using the precise definition
confirms this. 
\begin{align*}
\dims g'(x) & \dims=\lim_{h\rightarrow0}\frac{cf(x+h)-cf(x)}{h}\\
 & {\ec=c\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}}.
\end{align*}
 \begin{explanation}The constant multiple of a limit.\end{explanation}

\end{onlyenv}%{}

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

\noindent A close examination of the matter using the precise definition
confirms this. 
\begin{align*}
\dims g'(x) & \dims=c\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\
 & {\ec=cf'(x)}.
\end{align*}
 \begin{explanation}The precise definition of $f'(x)$.\end{explanation}

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

\lyxframeend{}\lyxframe{{Properties of derivatives}}

\noindent We have our {\hilite second} property:
\begin{theorem}%{}
\noindent [Properties of derivatives] Let $\ec f$ be a function. 
\begin{itemize}
\item If $\ec f(x)$ is constant, then the derivative of $\ec f$ is zero. 
\item If $\ec c$ is a constant and $\ec f'(x)$ exists,\\
 then the derivative of $\ec cf(x)$ is $\ec cf'(x)$. 
\end{itemize}
\end{theorem}%{}

\lyxframeend{}\lyxframe{{The derivative of a sum}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{onlyenv}%{}
<1>

\noindent Let $\ec f,g,h$ be functions such that $\ec s=f+g$. We
want to calculate $\ec s'(x)$; that is, the derivative of $\ec(f+g)(x)$.
From the definition of the derivative, 
\begin{align*}
\ec s'(x) & {\ec=\lim_{h\rightarrow0}\frac{s(x+h)-s(x)}{h}}.
\end{align*}


\end{onlyenv}%{}

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

\noindent Let $\ec f,g,h$ be functions such that $\ec s=f+g$. We
want to calculate $\ec s'(x)$; that is, the derivative of $\ec(f+g)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims s'(x) & \dims=\lim_{h\rightarrow0}\frac{s(x+h)-s(x)}{h}\\
 & {\ec=\lim_{h\rightarrow0}\frac{(f+g)(x+h)-(f+g)(x)}{h}}.
\end{align*}
 \begin{explanation}Definition of $s(x)$.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,h$ be functions such that $\ec s=f+g$. We
want to calculate $\ec s'(x)$; that is, the derivative of $\ec(f+g)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims s'(x) & \dims=\lim_{h\rightarrow0}\frac{(f+g)(x+h)-(f+g)(x)}{h}\\
 & {\ec=\lim_{h\rightarrow0}\frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}}.
\end{align*}
 \begin{explanation}Definition of addition of functions.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,h$ be functions such that $\ec s=f+g$. We
want to calculate $\ec s'(x)$; that is, the derivative of $\ec(f+g)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims s'(x) & \dims=\lim_{h\rightarrow0}\frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\\
 & {\ec=\lim_{h\rightarrow0}\left(\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}\right)}.
\end{align*}
 \begin{explanation}Group expressions with $f$, $g$.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,h$ be functions such that $\ec s=f+g$. We
want to calculate $\ec s'(x)$; that is, the derivative of $\ec(f+g)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims s'(x) & \dims=\lim_{h\rightarrow0}\left(\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}\right)\\
 & {\ec=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}+\lim_{h\rightarrow0}\frac{g(x+h)-g(x)}{h}}.
\end{align*}
 \begin{explanation}The limit of a sum is the sum of the limits.\\
 We are assuming that the limits being added exist.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,h$ be functions such that $\ec s=f+g$. We
want to calculate $\ec s'(x)$; that is, the derivative of $\ec(f+g)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims s'(x) & \dims=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}+\lim_{h\rightarrow0}\frac{g(x+h)-g(x)}{h}\\
 & \ec=f'(x)+g'(x).
\end{align*}
 \begin{explanation}The precise definition of the derivative.\end{explanation}

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

\lyxframeend{}\lyxframe{{Properties of derivatives}}

\noindent We have our {\hilite third} property: 
\begin{theorem}%{}
[Properties of derivatives]Let $\ec f$ be a function. 
\begin{itemize}
\item If $\ec f(x)$ is constant, then the derivative of $\ec f$ is zero. 
\item If $\ec c$ is a constant and $\ec f'(x)$ exists,\\
 then the derivative of $\ec cf(x)$ is $\ec cf'(x)$. 
\item If the derivatives of $\ec f(x)$ and $\ec g(x)$ exist,\\
 then the derivative of $\ec(f\pm g)(x)=f'(x)\pm g'(x)$. 
\end{itemize}
\end{theorem}%{}

\lyxframeend{}\lyxframe{{The derivative of a product}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{onlyenv}%{}
<1>

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
You might be inclinde to think that the derivative of a product is
the product of the derivatives; that is,
\[
\ec{(f(x)g(x))'=f'(x)g'(x)}.
\]
However, \emph{\hilite this does not work!} A little pencil grease
shows that the derivative of $\ec x^{2}$ is $\ec2x$, so the derivative
of $\ec x^{2}=x\cdot x$ cannot be $\ec x'\cdot x'=1\cdot1=1$. We
have to consider the definition of the derivative.

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\ec p'(x) & \ec=\lim_{h\rightarrow0}\frac{s(x+h)-s(x)}{h}
\end{align*}


\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims p'(x) & \dims=\lim_{h\rightarrow0}\frac{s(x+h)-s(x)}{h}\\
 & \ec=\lim_{h\rightarrow0}\frac{(fg)(x+h)-(fg)(x)}{h}
\end{align*}
 \begin{explanation}Definition of $\ec s(x)$.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims p'(x) & \dims=\lim_{h\rightarrow0}\frac{(fg)(x+h)-(fg)(x)}{h}\\
 & \ec=\lim_{h\rightarrow0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}
\end{align*}
 \begin{explanation}Definition of the product of two functions.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims p'(x) & \dims=\lim_{h\rightarrow0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\
 & \ec=\lim_{h\rightarrow0}\frac{f(x+h)g(x+h)+{\hilite\left[f(x+h)g(x)-f(x+h)g(x)\right]}-f(x)g(x)}{h}
\end{align*}
 \begin{explanation}Added zero: the terms in the middle cancel.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims p'(x) & \dims=\lim_{h\rightarrow0}\frac{f(x+h)g(x+h)+\left[f(x+h)g(x)-f(x+h)g(x)\right]-f(x)g(x)}{h}\\
 & \ec=\lim_{h\rightarrow0}\frac{f(x+h)\left[g(x+h)-g(x)\right]-g(x)\left[f(x+h)-f(x)\right]}{h}
\end{align*}
 \begin{explanation}Factored common terms.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims p'(x) & \dims=\lim_{h\rightarrow0}\frac{f(x+h)\left[g(x+h)-g(x)\right]-g(x)\left[f(x+h)-f(x)\right]}{h}\\
 & {\ec\overset{?}{=}\lim_{h\rightarrow0}\frac{f(x+h)\left[g(x+h)-g(x)\right]}{h}-\lim_{h\rightarrow0}\frac{g(x)\left[f(x+h)-f(x)\right]}{h}}.
\end{align*}
 \begin{explanation}Properties of limits, if these limits exist.\end{explanation}

\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\ec p'(x) & {\ec\overset{?}{=}\lim_{h\rightarrow0}\frac{f(x+h)\left[g(x+h)-g(x)\right]}{h}-\lim_{h\rightarrow0}\frac{g(x)\left[f(x+h)-f(x)\right]}{h}}.
\end{align*}
 We need to analyze the limits here. In the second case, $\ec g(x)$
is constant with respect to $\ec h$, so 
\[
{\ec\lim_{h\rightarrow0}\frac{g(x)\left[f(x+h)-f(x)\right]}{h}=g(x)\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=g(x)f'(x)}.
\]


\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\ec p'(x) & {\ec\overset{?}{=}\lim_{h\rightarrow0}\frac{f(x+h)\left(g(x+h)-g(x)\right]}{h}-\lim_{h\rightarrow0}\frac{g(x)\left[f(x+h)-f(x)\right]}{h}}.
\end{align*}
 In the first case, we use the properties of the limit of a product:
\[
{\ec\lim_{h\rightarrow0}\frac{f(x+h)\left[g(x+h)-g(x)\right]}{h}=\lim_{h\rightarrow0}f(x+h)\cdot\lim_{h\rightarrow0}\frac{g(x+h)-g(x)}{h}=f(x)\cdot g'(x)}.
\]


\end{onlyenv}%{}

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

\noindent Let $\ec f,g,p$ be functions such that $\ec p=fg$. We
want to calculate $\ec p'(x)$; that is, the derivative of $\ec(fg)(x)$.
From the definition of the derivative, 
\begin{align*}
\dims p'(x) & \dims=\lim_{h\rightarrow0}\frac{f(x+h)\left(g(x+h)-g(x)\right]}{h}-\lim_{h\rightarrow0}\frac{g(x)\left[f(x+h)-f(x)\right]}{h}\\
 & {\ec=f(x)g'(x)+f'(x)g(x)}.
\end{align*}
 Notice that we did \emph{not} obtain $\ec f'(x)g'(x)$.

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

\lyxframeend{}\lyxframe{{Properties of derivatives}}

\noindent We have our {\hilite fourth} property:
\begin{theorem}%{}
[Properties of derivatives]Let $\ec f$ be a function. 
\begin{itemize}
\item If $\ec f(x)$ is constant, then the derivative of $\ec f$ is zero. 
\item If $\ec c$ is a constant and $\ec f'(x)$ exists,\\
 then the derivative of $\ec cf(x)$ is $\ec cf'(x)$. 
\item If the derivatives of $\ec f(x)$ and $\ec g(x)$ exist,\\
 then the derivative of $\ec(f\pm g)(x)$ is $\ec f'(x)\pm g'(x)$. 
\item \emph{The product rule}\\
 If the derivatives of $\ec f(x)$ and $\ec g(x)$ exist,\\
 then the derivative of $\ec(fg)(x)$ is $\ec f(x)g'(x)+f'(x)g(x)$. 
\end{itemize}
\end{theorem}%{}

\lyxframeend{}\lyxframe{{The derivative of a quotient}}

\noindent We will not show all the details of the derivative of a
quotient, but the argument proceeds in much the same way as the argument
for the derivative of a product. (\emph{Try} it!) Instead, we list
all the properties on the next page.


\lyxframeend{}\lyxframe{{Properties of derivatives}}
\begin{theorem}%{}
[(Properties of derivatives)]Let $\ec f$ be a function. 
\begin{itemize}
\item If $\ec f(x)$ is constant, then the derivative of $\ec f$ is zero. 
\item If $\ec c$ is a constant and $\ec f'(x)$ exists,\\
 then the derivative of $\ec cf(x)$ is $\ec cf'(x)$. 
\item If the derivatives of $\ec f(x)$ and $\ec g(x)$ exist,\\
 then the derivative of $\ec(f\pm g)(x)$ is $\ec f'(x)\pm g'(x)$. 
\item \emph{(The product rule)}\\
 If the derivatives of $\ec f(x)$ and $\ec g(x)$ exist,\\
 then the derivative of $\ec(fg)(x)$ is $\ec f(x)g'(x)+f'(x)g(x)$. 
\item \emph{(The quotient rule)}\\
 If the derivatives of $\ec f(x)$ and $\ec g(x)$ exist,\\
 then the derivative of $\ec(f/g)(x)$ is $\ec{\displaystyle \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}$. 
\end{itemize}
\end{theorem}%{}

\lyxframeend{}\lyxframe{{Examples}}

Examples on the following slides use the facts that
\[
\mathrm{the\ derivative\ of\ }{\ec x^{2}+1}\mathrm{\ is\ }{\ec2x}\qquad\mathrm{\ and\ the\ derivative\ of\ }{\ec x^{3}-1}\mathrm{\ is\ }{\ec3x^{2}}.
\]
 You should verify these by hand, using the precise definition of
the derivative.


\lyxframeend{}\lyxframe{\noindent {Example}}

\noindent Using the {\hilite constant multiple rule}, the derivative
of $\ec4(x^{2}+1)$ is
\begin{align*}
\ec4\cdot\left[\frac{d}{dx}(x^{2}+1)\right] & {\ec=4\cdot2x}.
\end{align*}
 (One could simplify this further, but we want you to see the structure.)

\bigskip{}
You could also verify this by taking the derivative of the product
$\ec4(x^{2}+1)=4x^{2}+4$, but at this point it would be laborious
to use the precise derivative on the product.


\lyxframeend{}\lyxframe{\noindent {Example}}

\noindent Using the {\hilite product rule}, the derivative of $\ec(x^{2}+1)(x^{3}-1)$
is
\begin{align*}
\ec(x^{2}+1)\left[\frac{d}{dx}(x^{3}-1)\right]+\left[\frac{d}{dx}(x^{2}+1)\right](x^{3}-1) & {\ec=(x^{2}+1)\cdot3x^{2}+2x(x^{3}-1)}.
\end{align*}
 (One could simplify this further, but we want you to see the structure.)

\bigskip{}
You could also verify this by taking the derivative of the product
$\ec(x^{2}+1)(x^{3}-1)=x^{5}+x^{3}-x^{2}-1$ and comparing it to the
simplification of the derivative $\ec(x^{2}+1)\cdot3x^{2}+2x(x^{3}-1)=5x^{4}+3x^{2}-2x$,
but at this point it would be laborious to use the precise derivative
on the product.


\lyxframeend{}\lyxframe{\noindent {Example}}

\noindent Using the {\hilite quotient rule}, the derivative of
$\ec(x^{2}+1)/(x^{3}-1)$ is
\[
{\ec\frac{\left[\frac{d}{dx}(x^{2}+1)\right](x^{3}-1)-(x^{2}+1)\left[\frac{d}{dx}(x^{3}-1)\right]}{(x^{3}-1)^{2}}=\frac{2x(x^{3}-1)-(x^{2}+1)\cdot3x^{2}}{(x^{3}-1)^{2}}}.
\]
 (One could simplify this further, but we want you to see the structure.)


\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item We determined a definition for the derivative \emph{as a function
for arbitrary $\ec x$}:\setlength{\shadeboxwidth}{4in}

\begin{definition}%{}
[(The derivative as a function)]If a function $\ec f$ is differentiable
at every point in its domain, then the derivative of $\ec f$ is
\[
{\ec f'(x)=\lim_{x\rightarrow a}\frac{f(x+h)-f(x)}{h}}.
\]
 
\end{definition}%{}
\item We identified several properties of the derivative, for 

\begin{itemize}
\item a constant and a constant multiple of a function; 
\item a sum or difference of two functions; 
\item a product or quotitent of two functions.
\end{itemize}
\item The first three were rather intuitive, based on what we know of limits. 
\item The last two were counter-intuitive, and required us to play some
algebra in the precise definition of the derivative. 
\end{itemize}

\lyxframeend{}
\end{document}
