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
One-sided Limits
How can a limit fail to exist?
Infinite Limits and Vertical Asymptotes

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 Piece-wise Functions
The Squeeze Theorem

Continuity and the Intermediate Value Theorem

Definition of continuity
Continuity and piece-wise 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 $f'$
Notation and higher-order 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

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

How to tackle the problems
Example (ladder)
Example (shadow)

Linear Approximation and Differentials

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

The Extreme Value Theorem
Fermat's Theorem

The Mean Value and other Theorems

Rolle's Theorems
The Mean Value Theorem
Finding $c$

$f$ vs. $f'$

Increasing/Decreasing Test and Critical Numbers
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


Another Example

Newton's Method

The Idea of Newton's Method
An Example
Solving Transcendental Equations
When NM doesn't work


Anti-derivatives and Physics
Some formulas
Anti-derivatives are not Integrals

The Area under a curve

The Area Problem and Examples
Riemann Sums Notation

Definite Integrals

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


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)}{x-a}\quad}{\displaystyle\quad\frac{g(x)-g(a)}{x-a}\quad}} \cr\cr\cr\cr\cr &=& \displaystyle\frac{\displaystyle\lim_{x\to a} \,\,\Big(\frac{f(x)-f(a)}{x-a}\Big)\,\,}{\displaystyle\lim_{x\to a} \,\,\Big(\frac{g(x)-g(a)}{x-a}\Big)\,\,} \cr\cr\cr\cr\cr &=& \displaystyle{\frac{f'(a)}{g'(a)}}. \end{eqnarray*} By definition, $f'(a) = \displaystyle{\lim_{x\to a} \frac{f(x)-f(a)}{x-a}}$ and $g'(a) = \displaystyle{\lim_{x\to a}\frac{g(x)-g(a)}{x-a}}$. 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 half-open 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. $\qquad \hbox{QED}$