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%MODULE 1: Epsilon-Delta limits
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\begin_layout Title
A precise notion of the derivative
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\begin_layout Subtitle
In the previous module, you studied the 
\emph on
intuitive
\emph default
 idea of a derivative.
 In this module, you will learn 
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hilite
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 the 
\emph on
precise
\emph default
 definition of the derivative
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}
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.
 You will also see how the precise definition is related to limits.
 
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A Sample problem
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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.
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\begin_layout Standard
\noindent
He knows from physics that an object dropped above the earth obeys the function
\begin_inset Formula 
\[
\ec H(t)=-9.8t^{2}+h_{0}
\]

\end_inset

where
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec H(t)$
\end_inset

 is the height of the object above the ground, 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec t$
\end_inset

 is the time in seconds, and 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec h_{0}$
\end_inset

 is the height of the object at time 
\begin_inset Formula $\ec t=0$
\end_inset

.
 
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The roof of the Empire State Building is 1250
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ft above the ground.
 
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How fast is the superhero falling after five seconds?
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SAGE worksheets
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In this module you will not need a SAGE worksheet.
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\noindent
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{
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Precise definition of the derivative
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}
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\begin_layout Standard
\noindent
Recall from the previous module that the 
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 derivative
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 is 
\emph on
the instantaneous rate of change
\emph default
.
 We had described this as 
\emph on
the slope of the line tangent to
\emph default
 the curve.
 We further explained that this slope was 
\emph on
the limit of the slopes of the secant lines
\emph default
.
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\noindent
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{
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Precise definition of the derivative
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\begin_deeper
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\noindent
Let 
\begin_inset Formula $\ec f(x)$
\end_inset

 be a function, and 
\begin_inset Formula $\ec a$
\end_inset

 a real number.
 To find the derivative of 
\begin_inset Formula $\ec f$
\end_inset

 at 
\begin_inset Formula $\ec x=a$
\end_inset

, we need to find the limit of the slopes of the secant lines passing through
 
\begin_inset Formula $\ec x=a$
\end_inset

.
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\begin_layout Standard
That means the derivative of 
\begin_inset Formula $\ec f(x)$
\end_inset

 is
\begin_inset Formula 
\begin{align*}
\ec\lim_{x\rightarrow a}\frac{\Delta y}{\Delta x}.
\end{align*}

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\begin_layout Standard
\noindent
If we let 
\begin_inset Formula $\ec x$
\end_inset

 vary around 
\begin_inset Formula $\ec a$
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, then 
\begin_inset Formula $\ec\Delta x=x-a$
\end_inset

 and 
\begin_inset Formula $\ec\Delta y=f(x)-f(a)$
\end_inset

.
 That means the derivative of 
\begin_inset Formula $\ec f(x)$
\end_inset

 is
\begin_inset Formula 
\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*}

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\begin_layout Standard
\noindent
We usually write 
\begin_inset Formula $\ec f'(a)$
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 for the 
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 derivative
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 of 
\begin_inset Formula $\ec f(x)$
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 at 
\begin_inset Formula $\ec x=a$
\end_inset

, so
\begin_inset Formula 
\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*}

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Alternate definition
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\begin_layout Standard
Let 
\begin_inset Formula $f(x)$
\end_inset

 be a function, and 
\begin_inset Formula $a$
\end_inset

 a real number.
 To find the derivative of 
\begin_inset Formula $f$
\end_inset

 at 
\begin_inset Formula $x=a$
\end_inset

, we need to find the limit of the slopes of the secant lines passing through
 
\begin_inset Formula $x=a$
\end_inset

.
 
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sideways false
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\begin_layout Plain Layout
 
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	scale 20

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\begin_layout Standard
\noindent
A second definition considers 
\begin_inset Formula $\ec h=x-a$
\end_inset

.
 In this case the definition becomes
\begin_inset Formula 
\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*}

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Two definitions
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\begin_layout Standard
So there are two 
\begin_inset Quotes eld
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precise
\begin_inset Quotes erd
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 definitions of the derivative that you must keep in mind.
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\begin_layout Definition
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[
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(The derivative at a point)
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]
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The derivative of a function 
\begin_inset Formula $\ec f$
\end_inset

 at 
\begin_inset Formula $\ec x=a$
\end_inset

 is 
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\ec{\displaystyle f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}}$
\end_inset

, or equivalently 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec{\displaystyle f'(a)=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}}$
\end_inset

.
 
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\end_deeper
\begin_layout Standard
\noindent
You should think about why these two definitions are equivalent both algebraical
ly and geometrically.
 Re-examine the previous slides and compare the diagrams until you understand
 how they are the same.
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Two notations for the derivative
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\begin_deeper
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We usually write 
\begin_inset Formula $\ec y'$
\end_inset

 for the derivative of 
\begin_inset Formula $\ec y$
\end_inset

; this is called 
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 Newton's notation
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.
 We can also write 
\begin_inset Formula $\ec y''$
\end_inset

 for the derivative of the derivative, called the 
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 second derivative
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; 
\begin_inset Formula $\ec y'''$
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 for the derivative of the second derivative, called the 
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 third derivative
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; and so forth.
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\begin_layout Standard
\noindent
Another notation is 
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 Leibniz notation
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:
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\begin_inset Formula 
\begin{align*}
\ec y' & \ec\rightarrow\frac{dy}{dx}\\
\ec y'' & \ec\rightarrow\frac{d^{2}y}{dx^{2}}\\
 & \vdots
\end{align*}

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\begin_deeper
\begin_layout Standard
\noindent
This is related to the fact that
\begin_inset VSpace -1.5ex
\end_inset


\begin_inset Formula 
\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*}

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\begin_deeper
\begin_layout Standard
Do not confuse the notation 
\begin_inset Formula $\ec dy/dx$
\end_inset

 for a fraction of two quantities 
\begin_inset Formula $\ec dy$
\end_inset

 and 
\begin_inset Formula $\ec dx$
\end_inset

; it is simply a way of writing the derivative.
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An example
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\begin_deeper
\begin_layout Standard
We illustrate both definitions with the example given at the beginning of
 the module.
 We had 
\begin_inset Formula $\ec H(t)=-9.8t^{2}+h_{0}$
\end_inset

.
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\begin_deeper
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\noindent
Since the roof of the Empire State Building is 1250
\begin_inset space ~
\end_inset

ft above the ground, we can rewrite this as 
\begin_inset Formula 
\[
{\ec H(t)=-9.8t^{2}+1250}.
\]

\end_inset


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\begin_deeper
\begin_layout Standard
\noindent
Since the roof of the Empire State Building is 1250
\begin_inset space ~
\end_inset

ft above the ground, we can rewrite this as 
\begin_inset Formula 
\[
H(t)=-9.8t^{2}+1250.
\]

\end_inset

We want to know 
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How fast is the superhero falling after five seconds?
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\begin_deeper
\begin_layout Standard
\noindent
Since the roof of the Empire State Building is 1250
\begin_inset space ~
\end_inset

ft above the ground, we can rewrite this as 
\begin_inset Formula 
\[
H(t)=-9.8t^{2}+1250.
\]

\end_inset

We want to know 
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How fast is the superhero falling after five seconds?
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\begin_inset Quotes eld
\end_inset

How fast
\begin_inset Quotes erd
\end_inset

 is a 
\emph on
rate of change
\emph default
 of position.
 (In physics, we call this the 
\emph on
velocity
\emph default
.)
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\begin_deeper
\begin_layout Standard
\noindent
Since the roof of the Empire State Building is 1250
\begin_inset space ~
\end_inset

ft above the ground, we can rewrite this as 
\begin_inset Formula 
\[
H(t)=-9.8t^{2}+1250.
\]

\end_inset

We want to know 
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How fast is the superhero falling after five seconds?
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\end_inset

 
\begin_inset Quotes eld
\end_inset

How fast
\begin_inset Quotes erd
\end_inset

 is a 
\emph on
rate of change
\emph default
 of position.
 (In physics, we call this the 
\emph on
velocity
\emph default
.) So we need to compute the 
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 derivative
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 of 
\begin_inset Formula $\ec H(t)$
\end_inset

 at 
\begin_inset Formula $\ec t=5$
\end_inset

.
 
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\align center
That is, we need to compute 
\begin_inset Formula $\ec H'(5)$
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.
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\begin_inset Formula $H'(5)$
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: method 1
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Since we gave you two definitions of the 
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 derivative
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, we illustrate this computation two different ways.
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Using 
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 definition 1
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,
\begin_inset Formula 
\begin{align*}
\ec H'(5) & {\ec=\lim_{t\rightarrow5}\frac{H(t)-H(5)}{t-5}}.
\end{align*}

\end_inset

 We can't substitute 
\begin_inset Formula $\ec t=5$
\end_inset

, since that would give us zero in the denominator.
 We have to invest some effort into rewriting the fraction.
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 definition 1
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,
\begin_inset Formula 
\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*}

\end_inset

 Notice that 1250 cancels.
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 definition 1
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,
\begin_inset Formula 
\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*}

\end_inset

 The two terms of the numerator have a common factor,
\begin_inset space ~
\end_inset

9.8.
 Let's rewrite by factoring it out.
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 definition 1
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,
\begin_inset Formula 
\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*}

\end_inset

 Now recall from algebra that 
\begin_inset Formula $\ec t^{2}-25=(t-5)(t+5)$
\end_inset

.
 Let's rewrite it that way\SpecialChar \ldots{}

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 definition 1
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,
\begin_inset Formula 
\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*}

\end_inset

 \SpecialChar \ldots{}
allowing us to divide out the 
\begin_inset Formula $\ec t-5$
\end_inset

's that appear in the numerator and the denominator.
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 definition 1
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,
\begin_inset Formula 
\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*}

\end_inset

 We no longer have division by zero, so we can substitute 
\begin_inset Formula $\ec t=5$
\end_inset

 into the expression, and\SpecialChar \ldots{}

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 definition 1
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,
\begin_inset Formula 
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}{-9.8(t+5)}\\
\ec H'(5) & {\ec=-98}.
\end{align*}

\end_inset

 \SpecialChar \ldots{}
the superhero is falling at 98
\begin_inset space ~
\end_inset

ft/sec.
 (The negative indicates that he is falling.)
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 definition 1
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\end_inset

,
\begin_inset Formula 
\begin{align*}
\dims H'(5) & \dims=\lim_{t\rightarrow5}{-9.8(t+5)}\\
\ec H'(5) & {\ec=-98}.
\end{align*}

\end_inset

 \SpecialChar \ldots{}
the superhero is falling at 98
\begin_inset space ~
\end_inset

ft/sec.
 (The negative indicates that he is falling.)
\end_layout

\begin_layout Standard
\begin_inset VSpace bigskip
\end_inset

We could also write that 
\begin_inset Formula $\ec\frac{dH}{dt}=-98$
\end_inset

 (Leibniz notation).
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\begin_inset Formula $H'(5)$
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: method 2
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\noindent
Since we gave you two definitions of the 
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 derivative
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, we illustrate this computation two different ways.
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 definition 2
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\end_inset

,
\begin_inset Formula 
\begin{align*}
\ec H'(5) & {\ec=\lim_{h\rightarrow0}\frac{H(5+h)-H(5)}{h}}.
\end{align*}

\end_inset

 Again, we cannot substitute 
\begin_inset Formula $\ec h=0$
\end_inset

 to evaluate the limit, so we have to rewrite the fraction.
 We start by evaluating 
\begin_inset Formula $\ec H(5+h)$
\end_inset

 and 
\begin_inset Formula $\ec H(5)$
\end_inset

.
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 definition 2
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\end_inset

,
\begin_inset Formula 
\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*}

\end_inset

 Notice that 
\begin_inset Formula $\ec1250$
\end_inset

 cancels again this time.
 We'll simplify the numerator next.
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 definition 2
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\end_inset

,
\begin_inset Formula 
\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*}

\end_inset

 We need to distribute 
\begin_inset Formula $\ec-9.8$
\end_inset

.
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 definition 2
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,
\begin_inset Formula 
\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*}

\end_inset

 Again we have cancellation.
 When using this second definition of the derivative, 
\emph on
it is common for terms that do not contain 
\begin_inset Formula $\ec h$
\end_inset

 to cancel
\emph default
.
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 definition 2
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,
\begin_inset Formula 
\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*}

\end_inset

 Notice that 
\begin_inset Formula $\ec h$
\end_inset

 is a factor of every term in the numerator.
 Let's factor it out.
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 definition 2
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,
\begin_inset Formula 
\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*}

\end_inset

 This allows us to divide out 
\begin_inset Formula $\ec h$
\end_inset

, eliminating it from the denominator.
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 definition 2
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,
\begin_inset Formula 
\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*}

\end_inset

 We can now substitute 
\begin_inset Formula $\ec h=0$
\end_inset

\SpecialChar \ldots{}

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 definition 2
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,
\begin_inset Formula 
\begin{align*}
\dims H'(5) & \dims=\lim_{h\rightarrow0}(-98-\cancelto{0}{9.8h})\\
\ec H'(5) & {\ec=-98}.
\end{align*}

\end_inset

 \SpecialChar \ldots{}
we obtain the same answer as before: the superhero is falling at 98
\begin_inset space ~
\end_inset

ft/sec.
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Algebra
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\begin_layout Standard
You will have noticed that it was necessary to perform 
\emph on
some
\emph default
 sort of algebraic manipulation whenever applying the precise definition
 of the derivative.
 This is usually the case.
 Techniques we used in this example were: 
\end_layout

\begin_layout Itemize
distribution, 
\end_layout

\begin_layout Itemize
factoring, and 
\end_layout

\begin_layout Itemize
dividing common factors in fractions.
 
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
 You will also find it necessary to use other algebraic techniques, such
 as 
\end_layout

\begin_layout Itemize
rationalizing the denominator (multiply by conjugate), and 
\end_layout

\begin_layout Itemize
simplifying complex fractions (multiply numerator by reciprocal of denominator).
 
\end_layout

\begin_layout Standard
\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.
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Conclusion
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\begin_layout Standard
You have now learned two different, but equivalent definitions of the derivative.
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\begin_layout Definition
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[
\end_layout

\end_inset

(The derivative at a point)
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]
\end_layout

\end_inset

 The derivative of a function 
\begin_inset Formula $\ec f$
\end_inset

 at 
\begin_inset Formula $\ec x=a$
\end_inset

 is 
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\ec{\displaystyle f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}}$
\end_inset

, or equivalently 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec{\displaystyle f'(a)=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}}$
\end_inset

.
 
\end_layout

\end_deeper
\begin_layout Standard
\noindent
You should memorize them, but more importantly you should understand 
\emph on
why
\emph default
 the definition is defined with them.
 You will use them to compute many derivatives in the coming days.
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\end_body
\end_document
