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\begin{document}
%MODULE 1: Epsilon-Delta limits


%USER DEFINED COMMANDS
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%INPUT THE INSTRUCTIONS PAGE 



\title{A precise notion of the derivative}

\makebeamertitle
%BUILD  TITLE PAGE 



\subtitle{In the previous module, you studied the \emph{intuitive} idea of
a derivative. In this module, you will learn {\hilite the \emph{precise}
definition of the derivative}. You will also see how the precise
definition is related to limits. }


\lyxframeend{}\lyxframe{{A Sample problem}}

A supervillain throws a superhero off the roof of the Empire State
Building in New York. This particular superhero cannot fly, but he
can bounce off the ground. Having nothing in particular to think about
as he falls, he wonders how fast he travels at any given moment.

\noindent He knows from physics that an object dropped above the earth
obeys the function
\[
\ec H(t)=-9.8t^{2}+h_{0}
\]
where
\begin{itemize}
\item $\ec H(t)$ is the height of the object above the ground, 
\item $\ec t$ is the time in seconds, and 
\item $\ec h_{0}$ is the height of the object at time $\ec t=0$. 
\end{itemize}
The roof of the Empire State Building is 1250~ft above the ground.
\begin{question}How fast is the superhero falling after five seconds?\end{question}


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

In this module you will not need a SAGE worksheet.


\lyxframeend{}\lyxframe{\noindent {Precise definition of the derivative}}

\noindent Recall from the previous module that the {\hilite derivative}
is \emph{the instantaneous rate of change}. We had described this
as \emph{the slope of the line tangent to} the curve. We further explained
that this slope was \emph{the limit of the slopes of the secant lines}.


\lyxframeend{}\lyxframe{\noindent {Precise definition of the derivative}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent Let $\ec f(x)$ be a function, and $\ec a$ a real number.
To find the derivative of $\ec f$ at $\ec x=a$, we need to find
the limit of the slopes of the secant lines passing through $\ec x=a$.
\begin{onlyenv}%{}
<1>

\noindent 
\begin{figure}[h!]
\centering{}\includegraphics[scale=0.2]{figures/basic_f} 
\end{figure}


\end{onlyenv}%{}

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

\noindent 
\begin{figure}[h!]
 

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



That means the derivative of $\ec f(x)$ is
\begin{align*}
\ec\lim_{x\rightarrow a}\frac{\Delta y}{\Delta x}.
\end{align*}


\end{onlyenv}%{}

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

\noindent 
\begin{figure}[h!]
 

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



\noindent If we let $\ec x$ vary around $\ec a$, then $\ec\Delta x=x-a$
and $\ec\Delta y=f(x)-f(a)$. That means the derivative of $\ec f(x)$
is
\begin{align*}
\ec\lim_{x\rightarrow a}\frac{\Delta y}{\Delta x} & {\ec=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}}.
\end{align*}


\end{onlyenv}%{}

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

\begin{figure}[h!]
 

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



\noindent We usually write $\ec f'(a)$ for the {\hilite derivative}
of $\ec f(x)$ at $\ec x=a$, so
\begin{align*}
\ec f'(a) & \ec=\lim_{x\rightarrow a}\frac{\Delta y}{\Delta x}\\
\ec & {\ec=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}}.
\end{align*}


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

\lyxframeend{}\lyxframe{{Alternate definition}}

Let $f(x)$ be a function, and $a$ a real number. To find the derivative
of $f$ at $x=a$, we need to find the limit of the slopes of the
secant lines passing through $x=a$. 
\begin{figure}[h!]
 

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


\noindent A second definition considers $\ec h=x-a$. In this case
the definition becomes
\begin{align*}
\ec f'(a) & \ec=\lim_{x\rightarrow a}\frac{\Delta y}{\Delta x}\\
 & {\ec f'(a)=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}}.
\end{align*}



\lyxframeend{}\lyxframe{{Two definitions}}

So there are two {}``precise'' definitions of the derivative that
you must keep in mind.
\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}%{}
\noindent You should think about why these two definitions are equivalent
both algebraically and geometrically. Re-examine the previous slides
and compare the diagrams until you understand how they are the same.


\lyxframeend{}\lyxframe{{Two notations for the derivative}}
\begin{overlayarea}
{\textwidth}{6in}

We usually write $\ec y'$ for the derivative of $\ec y$; this is
called {\hilite Newton's notation}. We can also write $\ec y''$
for the derivative of the derivative, called the {\hilite second
derivative}; $\ec y'''$ for the derivative of the second derivative,
called the {\hilite third derivative}; and so forth.


\noindent Another notation is {\hilite Leibniz notation}:\vspace{-1.5ex}
\begin{align*}
\ec y' & \ec\rightarrow\frac{dy}{dx}\\
\ec y'' & \ec\rightarrow\frac{d^{2}y}{dx^{2}}\\
 & \vdots
\end{align*}

\begin{onlyenv}%{}
<1>

\noindent This is related to the fact that\vspace{-1.5ex}
\begin{align*}
\ec y' & \ec=\lim_{\Delta x\rightarrow0}\frac{\Delta y}{\Delta x}\\
\ec y'' & \ec=\lim_{\Delta x\rightarrow0}\frac{\Delta y'}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{\Delta\left(\lim_{\Delta x\rightarrow0}\frac{\Delta y}{\Delta x}\right)}{\Delta x}\\
 & \vdots
\end{align*}


\end{onlyenv}%{}

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

Do not confuse the notation $\ec dy/dx$ for a fraction of two quantities
$\ec dy$ and $\ec dx$; it is simply a way of writing the derivative.

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

\lyxframeend{}\lyxframe{{An example}}
\begin{overlayarea}
{\textwidth}{6in}

We illustrate both definitions with the example given at the beginning
of the module. We had $\ec H(t)=-9.8t^{2}+h_{0}$.


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

\noindent Since the roof of the Empire State Building is 1250~ft
above the ground, we can rewrite this as 
\[
{\ec H(t)=-9.8t^{2}+1250}.
\]


\end{onlyenv}%{}

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

\noindent Since the roof of the Empire State Building is 1250~ft
above the ground, we can rewrite this as 
\[
H(t)=-9.8t^{2}+1250.
\]
We want to know \begin{question}How fast is the superhero falling
after five seconds?\end{question}

\end{onlyenv}%{}

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

\noindent Since the roof of the Empire State Building is 1250~ft
above the ground, we can rewrite this as 
\[
H(t)=-9.8t^{2}+1250.
\]
We want to know \begin{question}How fast is the superhero falling
after five seconds?\end{question} {}``How fast'' is a \emph{rate
of change} of position. (In physics, we call this the \emph{velocity}.)

\end{onlyenv}%{}

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

\noindent Since the roof of the Empire State Building is 1250~ft
above the ground, we can rewrite this as 
\[
H(t)=-9.8t^{2}+1250.
\]
We want to know \begin{question}How fast is the superhero falling
after five seconds?\end{question} {}``How fast'' is a \emph{rate
of change} of position. (In physics, we call this the \emph{velocity}.)
So we need to compute the {\hilite derivative} of $\ec H(t)$ at
$\ec t=5$. 

\end{onlyenv}%{}

\begin{center}
That is, we need to compute $\ec H'(5)$.
\par\end{center}

\end{overlayarea}

\lyxframeend{}\lyxframe{{$H'(5)$: method 1}}
\begin{overlayarea}
{\textwidth}{6in}

Since we gave you two definitions of the {\hilite derivative},
we illustrate this computation two different ways.
\begin{onlyenv}%{}
<1>

Using {\hilite definition 1},
\begin{align*}
\ec H'(5) & {\ec=\lim_{t\rightarrow5}\frac{H(t)-H(5)}{t-5}}.
\end{align*}
 We can't substitute $\ec t=5$, since that would give us zero in
the denominator. We have to invest some effort into rewriting the
fraction.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 1},
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}\frac{H(t)-H(5)}{t-5}\\
\ec H'(5) & {\ec=\lim_{t\rightarrow5}\frac{(-9.8t^{2}+\cancel{1250})-(-9.8(5)^{2}+\cancel{1250})}{t-5}}.
\end{align*}
 Notice that 1250 cancels.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 1},
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}\frac{(-9.8t^{2}+\cancel{1250})-(-9.8(5)^{2}+\cancel{1250})}{t-5}\\
\ec H'(5) & {\ec=\lim_{t\rightarrow5}\frac{-9.8t^{2}+9.8(5)^{2}}{t-5}}.
\end{align*}
 The two terms of the numerator have a common factor,~9.8. Let's
rewrite by factoring it out.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 1},
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}\frac{-9.8t^{2}+9.8(5)^{2}}{t-5}\\
\ec H'(5) & {\ec=\lim_{t\rightarrow5}\frac{-9.8(t^{2}-25)}{t-5}}.
\end{align*}
 Now recall from algebra that $\ec t^{2}-25=(t-5)(t+5)$. Let's rewrite
it that way\ldots{}

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 1},
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}\frac{-9.8(t^{2}-25)}{t-5}\\
\ec H'(5) & {\ec=\lim_{t\rightarrow5}\frac{(-9.8(t+5)\cancel{(t-5)}}{\cancel{t-5}}}.
\end{align*}
 \ldots{}allowing us to divide out the $\ec t-5$'s that appear in
the numerator and the denominator.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 1},
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}\frac{(-9.8(t+5)\cancel{(t-5)}}{\cancel{t-5}}\\
\ec H'(5) & {\ec=\lim_{t\rightarrow5}{-9.8(t+5)}}.
\end{align*}
 We no longer have division by zero, so we can substitute $\ec t=5$
into the expression, and\ldots{}

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 1},
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}{-9.8(t+5)}\\
\ec H'(5) & {\ec=-98}.
\end{align*}
 \ldots{}the superhero is falling at 98~ft/sec. (The negative indicates
that he is falling.)

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 1},
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}{-9.8(t+5)}\\
\ec H'(5) & {\ec=-98}.
\end{align*}
 \ldots{}the superhero is falling at 98~ft/sec. (The negative indicates
that he is falling.)


\bigskip{}
We could also write that $\ec\frac{dH}{dt}=-98$ (Leibniz notation).

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

\lyxframeend{}\lyxframe{\noindent {$H'(5)$: method 2}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent Since we gave you two definitions of the {\hilite derivative},
we illustrate this computation two different ways.
\begin{onlyenv}%{}
<1>

\noindent Using {\hilite definition 2},
\begin{align*}
\ec H'(5) & {\ec=\lim_{h\rightarrow0}\frac{H(5+h)-H(5)}{h}}.
\end{align*}
 Again, we cannot substitute $\ec h=0$ to evaluate the limit, so
we have to rewrite the fraction. We start by evaluating $\ec H(5+h)$
and $\ec H(5)$.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 2},
\begin{align*}
\dims H'(5) & \dims=\lim_{h\rightarrow0}\frac{H(5+h)-H(5)}{h}\\
\ec H'(5) & {\ec=\lim_{h\rightarrow0}\frac{(-9.8(5+h)^{2}+\cancel{1250})-(-9.8\cdot5^{2}+\cancel{1250})}{h}}.
\end{align*}
 Notice that $\ec1250$ cancels again this time. We'll simplify the
numerator next.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 2},
\begin{align*}
\dims H'(5) & \dims=\lim_{h\rightarrow0}\frac{(-9.8(5+h)^{2}+\cancel{1250})-(-9.8\cdot5^{2}+\cancel{1250})}{h}\\
\ec H'(t) & {\ec=\lim_{h\rightarrow0}\frac{-9.8(25+10h+h^{2})+9.8\cdot25}{h}}.
\end{align*}
 We need to distribute $\ec-9.8$.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 2},
\begin{align*}
\dims H'(t) & \dims=\lim_{h\rightarrow0}\frac{-9.8(25+10h+h^{2})+9.8\cdot25}{h}\\
\ec H'(t) & {\ec=\lim_{h\rightarrow0}\frac{\cancel{-9.8\cdot25}-98h-9.8h^{2}+\cancel{9.8\cdot25}}{h}}.
\end{align*}
 Again we have cancellation. When using this second definition of
the derivative, \emph{it is common for terms that do not contain $\ec h$
to cancel}.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 2},
\begin{align*}
\dims H'(t) & \dims=\lim_{h\rightarrow0}\frac{\cancel{-9.8\cdot25}-98h-9.8h^{2}+\cancel{9.8\cdot25}}{h}\\
\ec H'(t) & {\ec=\lim_{h\rightarrow0}\frac{-98h-9.8h^{2}}{h}}.
\end{align*}
 Notice that $\ec h$ is a factor of every term in the numerator.
Let's factor it out.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 2},
\begin{align*}
\dims H'(t) & \dims=\lim_{h\rightarrow0}\frac{-98h-9.8h^{2}}{h}\\
\ec H'(t) & {\ec=\lim_{h\rightarrow0}\frac{\cancel{h}(-98-9.8h)}{\cancel{h}}}.
\end{align*}
 This allows us to divide out $\ec h$, eliminating it from the denominator.

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 2},
\begin{align*}
\dims H'(t) & \dims=\lim_{h\rightarrow0}\frac{\cancel{h}(-98-9.8h)}{\cancel{h}}\\
\ec H'(5) & {\ec=\lim_{h\rightarrow0}(-98-\cancelto{0}{9.8h})}.
\end{align*}
 We can now substitute $\ec h=0$\ldots{}

\end{onlyenv}%{}

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

\noindent Using {\hilite definition 2},
\begin{align*}
\dims H'(5) & \dims=\lim_{h\rightarrow0}(-98-\cancelto{0}{9.8h})\\
\ec H'(5) & {\ec=-98}.
\end{align*}
 \ldots{}we obtain the same answer as before: the superhero is falling
at 98~ft/sec.

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

\lyxframeend{}\lyxframe{{Algebra}}

You will have noticed that it was necessary to perform \emph{some}
sort of algebraic manipulation whenever applying the precise definition
of the derivative. This is usually the case. Techniques we used in
this example were: 
\begin{itemize}
\item distribution, 
\item factoring, and 
\item dividing common factors in fractions. 
\end{itemize}

\pause{}

 You will also find it necessary to use other algebraic techniques,
such as 
\begin{itemize}
\item rationalizing the denominator (multiply by conjugate), and 
\item simplifying complex fractions (multiply numerator by reciprocal of
denominator). 
\end{itemize}
\noindent You will also need to use techniques for taking limits of
trigonometric functions. This is covered in your text, and your instructor
will also discuss it.


\lyxframeend{}\lyxframe{{Conclusion}}

You have now learned two different, but equivalent definitions of
the derivative.
\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}%{}
\noindent You should memorize them, but more importantly you should
understand \emph{why} the definition is defined with them. You will
use them to compute many derivatives in the coming days.


\lyxframeend{}
\end{document}
