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The Six Pillars of Calculus
The Pillars: A Road Map
A picture is worth 1000 words
Trigonometry Review
The basic trig functions
Basic trig identities
The unit circle
Addition of angles, double and half angle formulas
The law of sines and the law of cosines
Graphs of Trig Functions
Exponential Functions
Exponentials with positive integer exponents
Fractional and negative powers
The function $f(x)=a^x$ and its graph
Exponential growth and decay
Logarithms and Inverse functions
Inverse Functions
How to find a formula for an inverse function
Logarithms as Inverse Exponentials
Inverse Trig Functions
Intro to Limits
Close is good enough
Definition
Onesided Limits
How can a limit fail to exist?
Infinite Limits and Vertical Asymptotes
Summary
Limit Laws and Computations
A summary of Limit Laws
Why do these laws work?
Two limit theorems
How to algebraically manipulate a 0/0?
Limits with fractions
Limits with Absolute Values
Limits involving Rationalization
Limits of Piecewise Functions
The Squeeze Theorem
Continuity and the Intermediate Value Theorem
Definition of continuity
Continuity and piecewise functions
Continuity properties
Types of discontinuities
The Intermediate Value Theorem
Examples of continuous functions
Limits at Infinity
Limits at infinity and horizontal asymptotes
Limits at infinity of rational functions
Which functions grow the fastest?
Vertical asymptotes (Redux)
Toolbox of graphs
Rates of Change
Tracking change
Average and instantaneous velocity
Instantaneous rate of change of any function
Finding tangent line equations
Definition of derivative
The Derivative Function
The derivative function
Sketching the graph of $f'$
Differentiability
Notation and higherorder derivatives
Basic Differentiation Rules
The Power Rule and other basic rules
The derivative of $e^x$
Product and Quotient Rules
The Product Rule
The Quotient Rule
Derivatives of Trig Functions
Two important Limits
Sine and Cosine
Tangent, Cotangent, Secant, and Cosecant
Summary
The Chain Rule
Two forms of the chain rule
Version 1
Version 2
Why does it work?
A hybrid chain rule
Implicit Differentiation
Introduction and Examples
Derivatives of Inverse Trigs via Implicit Differentiation
A Summary
Derivatives of Logs
Formulas and Examples
Logarithmic Differentiation
Derivatives in Science
In Physics
In Economics
In Biology
Related Rates
Overview
How to tackle the problems
Example (ladder)
Example (shadow)
Linear Approximation and Differentials
Overview
Examples
An example with negative $dx$
Differentiation Review
Basic Building Blocks
Advanced Building Blocks
Product and Quotient Rules
The Chain Rule
Combining Rules
Implicit Differentiation
Logarithmic Differentiation
Conclusions and Tidbits
Absolute and Local Extrema
Definitions
The Extreme Value Theorem
Fermat's Theorem
Howto
The Mean Value and other Theorems
Rolle's Theorems
The Mean Value Theorem
Finding $c$
$f$ vs. $f'$
Increasing/Decreasing Test and Critical Numbers
Howto
The First Derivative Test
Concavity, Points of Inflection, and the Second Derivative Test
Indeterminate Forms and L'Hospital's Rule
What does $\frac{0}{0}$ equal?
Indeterminate Differences
Indeterminate Powers
Three Versions of L'Hospital's Rule
Proofs
Optimization
Strategies
Another Example
Newton's Method
The Idea of Newton's Method
An Example
Solving Transcendental Equations
When NM doesn't work
Antiderivatives
Antiderivatives and Physics
Some formulas
Antiderivatives are not Integrals
The Area under a curve
The Area Problem and Examples
Riemann Sums Notation
Summary
Definite Integrals
Definition
Properties
What is integration good for?
More Examples
The Fundamental Theorem of Calculus
Three Different Quantities
The Whole as Sum of Partial Changes
The Indefinite Integral as Antiderivative
The FTC and the Chain Rule


Proofs
Proof of Baby L'Hospital's Rule:
Suppose that $f(a)=g(a)=0$ and $g'(a) \ne 0$. Then, for any
$x$, $f(x)=f(x)f(a)$ and $g(x)=g(x)g(a)$. But
then,\begin{eqnarray*}
\displaystyle{\lim_{x \to a} \frac{f(x)}{g(x)}} &=&
\displaystyle{\lim_{x \to a} \,\frac{f(x)f(a)}{g(x)g(a)}}\cr\cr\cr\cr\cr
&=& \displaystyle{ \lim_{x\to a}\,
\frac{\displaystyle\quad\frac{f(x)f(a)}{xa}\quad}{\displaystyle\quad\frac{g(x)g(a)}{xa}\quad}} \cr\cr\cr\cr\cr
&=&
\displaystyle\frac{\displaystyle\lim_{x\to a} \,\,\Big(\frac{f(x)f(a)}{xa}\Big)\,\,}{\displaystyle\lim_{x\to a} \,\,\Big(\frac{g(x)g(a)}{xa}\Big)\,\,} \cr\cr\cr\cr\cr
&=& \displaystyle{\frac{f'(a)}{g'(a)}}. \end{eqnarray*}
By definition,
$f'(a) = \displaystyle{\lim_{x\to a} \frac{f(x)f(a)}{xa}}$ and
$g'(a) = \displaystyle{\lim_{x\to a}\frac{g(x)g(a)}{xa}}$.
Since $f'$ and $g'$ are assumed to be continuous, $f'(a)/g'(a)$ is also
$$\frac{\lim_{x \to a} f'(x)}{\lim_{x\to a}g'(x)} =
\lim_{x\to a}\frac{f'(x)}{g'(x)}.\qquad \hbox{QED}$$

To prove Macho L'Hospital's Rule we first need a lemma:
Souped Up Mean Value Theorem: If $f(x)$
and $g(x)$ are continuous on a closed interval $[a,b]$ and
differentiable on the open interval $(a,b)$, then there is a
point $c$, between $a$ and $b$, where
$$\big(f(b)f(a)\big)g'(c) = \big(g(b)g(a)\big)f'(c). $$
(When $g(x)=x$,
this is the same as the usual MVT.)
Proof: Consider the
function $$ h(x) = \big(f(x)f(a)\big)\big(g(b)g(a)\big)  \big(f(b)f(a)\big)\big(g(x)g(a)\big).$$
This is continuous on $[a,b]$ and differentiable on $(a,b)$, with
$$h'(x) = f'(x)\big(g(b)g(a)\big)  g'(x) \big(f(b)  f(a)\big).$$ Note that
$h(a)=0=h(b)$. By Rolle's Theorem, there a spot $c$ where
$h'(c)=0$. But $h'(c)= \big(f(b)f(a)\big)g'(c) 
\big(g(b)g(a)\big)f'(c).$ Since this is zero, $\big(f(b)f(a)\big)g'(c) = \big(g(b)g(a)\big)f'(c)$. $\qquad \hbox{QED}$

Proof of Macho L'Hospital's Rule:
By assumption,
$f$ and $g$ are differentiable to the right of $a$, and the limits of
$f$ and $g$ as $x \to a^+$ are zero. Define $f(a)$ to be zero, and
likewise define $g(a)=0$. Since these values agree with the limits,
$f$ and $g$ are continuous on some halfopen interval $[a,b)$ and
differentiable on$(a,b)$.
For any $x \in (a,b)$, we have that $f$ and $g$ are
differentiable on $(a,x)$ and continuous on $[a,x]$. By the Souped up
MVT, there is apoint $c$ between $a$ and $x$ such that $f'(c) g(x) =
f'(x) g(c)$. In other words, $f'(c)/g'(c) = f(x)/g(x)$. Also, as
$x$ approaches $a$, $c$ also approaches $a$, since $c$ is somewhere
between $x$ and $a$. But then
$$\lim_{x \to a^+}\, \frac{f(x)}{g(x)} =
\lim_{x \to a^+} \,\frac{f'(c)}{g'(c)} = \lim_{c \to
a^+}\,\frac{f'(c)}{g'(c)}.$$
That last expression is the same as $\lim_{x \to a^+} f'(x)/g'(x)$. $\qquad \hbox{QED}$ 
Proof of the Extended L'Hospital's Rule:
We're going to use a single trick, over and over again. Namely, we
can always rewrite $x$ as $\frac{1}{1/x}$, $f(x)$ as $\frac{1}{1/f(x)}$ and $g(x)$
as $\frac{1}{1/g(x)}$.
Suppose $L= \lim_{x \to a} \frac{f(x)}{g(x)}$, where both $f$
and $g$ go to $\infty$ (or $\infty$) as $x \to a$. Also suppose that
$L$ is neither 0 nor infinite. Then $$ L = \lim_{x \to a}\,
\frac{f(x)}{g(x)} = \lim_{x \to a}\, \frac{1/g(x)}{1/f(x)}.$$ Since
$1/g(x)$ and $1/f(x)$ go to zero as $x \to a$, we can apply the (baby
or macho) L'Hospital's rule to this limit:
\begin{eqnarray*}L &=& \lim_{x \to a}\, \frac{ (1/g)'}{(1/f)'}
\cr\cr\cr&=& \lim_{x \to a} \,\frac{g'(x)/g(x)^2}{f'(x)/f(x)^2}
\cr\cr\cr&=& \lim_{x\to a}\, \frac{f(x)^2 \cdot g'(x)}{g(x)^2\cdot f'(x)}
\cr\cr\cr&=& \lim_{x \to a}\,\frac{f(x)^2}{g(x)^2}\,\cdot\, \lim_{x \to
a}\, \frac{g'(x)}{f'(x)} \cr\cr\cr& = & \frac{L^2}{\displaystyle\lim_{x \to
a}\, \Big(f'(x)/g'(x)\Big)}.\end{eqnarray*}
Since $L = \displaystyle\frac{L^2}{\lim_{x \to a} \big(f'(x)/g'(x)\big)}$, $L$ must equal
$\displaystyle \lim_{x \to a}\Big(f'(x)/g'(x)\Big)$, which is what we wanted to
prove.
This argument only works for finite and nonzero values of
$L$. However, if $L=0$, we can apply the same argument to the limit
of $\big(f(x)+g(x)\big)/g(x)$, which then does not equal zero. The upshot
is that
$$ 1 + \lim_{x \to a}\,\frac{f(x)}{g(x)}
= \lim_{x\to a}\, \frac{f(x)+g(x)}{g(x)}
= \lim_{x \to a}\, \frac{f'(x) + g'(x)}{g'(x)}
= 1 + \lim_{x \to a}\,\frac{f'(x)}{g'(x)},$$
hence that $\lim\, (f/g) = \lim \,(f'/g')$. Finally, if $\lim\,(f/g)=\pm
\infty$, look instead at $\lim\,(g/f)$, which is then zero, so the
previous reasoning applies. Since $0=\lim\,(g/f)=\lim\,(g'/f')$,
$\lim\,(f'/g')$ must be infinite. By the Souped up MVT, $f/g$ has
the same sign as $f'/g'$, so we must have $\lim\,(f/g)=\lim\,(f'/g')$.
Now that we have L'Hopital's Rule for limits as $x \to a$ (or $x
\to a^+$ or $x \to a^$), we consider what happens as $x \to
\infty$. Define a new variable $t = 1/x$, so that $x \to \infty$ is
the same as $t \to 0^+$.Then $$ \lim_{x \to \infty}\, \frac{f(x)}{g(x)}
= \lim_{t \to 0^+}\, \frac{f(1/t)}{g(1/t)}.$$ But we know how to apply
L'Hospital's Rule to limits as $t \to 0$, so this turns
into
$$\lim_{t \to 0^+}\, \frac{\frac{d}{dt}
f(1/t)}{\frac{d}{dt}g(1/t)} = \lim_{t \to 0^+}
\frac{f'(1/t)\cdot \frac{1}{t^2}}{g'(1/t)\cdot \frac{1}{t^2}}= \lim_{t \to 0^+}
\frac{f'(1/t)}{g'(1/t)}.$$
Converting back to $x=1/t$, we get $$
\lim_{x \to \infty} \frac{f'(x)}{g'(x)},$$ which is what we
wanted.
Computing a limit as $x \to \infty$ is similar, only with $t \to
0^$ instead of $t \to 0^+$. That completes the proof of
L'Hospital's Rule. $\qquad \hbox{QED}$ 
