Calculus Index Cards

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One of the most common student questions is:

‘What do I need to know?’

Answer:   the ‘Index Card Method’

When I teach, every concept becomes an index card (4" x 6", blank both sides—it's easier to write math without lines). The cards guide class discussions.

Students hand-copy the cards (adding additional information, as needed). By the end of the course, they have a stack that embodies every key idea. Cards can be mixed up so learning isn't order-dependent; ones that are well understood are taken out to focus on those that remain.

Students hold the whole course in the palm of their hand.

I used this method for decades of teaching. Early on, the classroom blackboard became ‘one side of a card’: as I lectured, the card contents emerged on the board, and students took notes. Later on, students got photo-copies of cards, or they were posted on the web. Many-a-student has told me (years later) that they still have their stack of cards!

These are not typical ‘flash cards’, which might have (say) a theorem name on one side and statement on the other. Instead, they have enough meat to form the basis for a complete calculus course—just not much fat or gravy! They've been fine-tuned: with only 4" x 6" of space, what is the essential content to include, and what's the best way to say it?

I've included links to my own online materials. For my online Calculus course, you can omit the ‘included for higher-level course’ cards. To clarify handwriting and abbreviations, there's a ‘transcript’ button above each image. (I hope these transcripts are useful, since they represent about 70 hours of typesetting work!)

1.

Average Rate of Change
Tangent Lines

AVERAGE RATE OF CHANGE
... of a function $\,f\,$ between $\,a\,$ and $\,b\,$ is: $$\frac{f(b)-f(a)}{b-a} = \frac{\text{change in output}}{\text{change in input}}$$ It is the slope of the secant line thru $\,(a,f(a))\,$ and $\,(b,f(b))\,$
AVROC has units: $\,\frac{\text{units of $\,f(x)\,$}}{\text{units of $\,x\,$}}$
[labels on graph] $\,a\,,$ $\,b\,,$ $\,f(a)\,,$ $\,f(b)\,,$ $\,y = f(x)\,$

THE TANGENT PROBLEM
‘Tangent’ derives from the Latin ‘tangens’, meaning ‘touching’.
The tangent line to a curve at a point (when it exists) captures the direction you're moving at the instant you pass thru this point.
Strategy: Choose a nearby pt $\,(x,x^2)\,$; $$ \begin{align} &\text{slope of line between $\,(3,9)\,$ and $\,(x,x^2)\,$}\cr &\ \ = \frac{x^2-9}{x-3} \\ &\ \ = \frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}} \\ &\ \ = x + 3\,, \text{for $\,x \ne 3\,$} \end{align} $$ [labels on graph] $\,3\,,$ $\,x\,,$ $\,9\,,$ $\,x^2\,,$ $\,y = x^2\,$
[arrow pointing to line] What is this slope?
A tangent line gives an instantaneous rate of change.

2.

Analyzing a Falling Object

POSITION FUNCTIONS
Spse [suppose] you live on a long, straight road.
Your house is at position $\,0\,.$ You leave at $\,t = 0\,$ and drive in the direction shown.

• drive at a constant speed until $\,t = a\,$
• drive at a faster constant speed until $\,t = b\,$
• gradually slow down
• come to rest @ $\,t = c\,$

[labels on graph] $\,a\,,$ $\,b\,,$ $\,c\,,$ $\,t\,,$ $\,y\,,$ $\,y = p(t)\,$

the DERIVATIVE OF A FUNCTION ...
... gives the slopes of the tangent lines (when they exist)
$\,\frac{\Delta \text{position}}{\Delta \text{time}} = \text{velocity}$; the derivative of a position function is a velocity fct [function]
$\,\frac{\Delta \text{velocity}}{\Delta \text{time}} = \text{acceleration}$; the derivative of a velocity function is an acceleration fct
Note: $\,\text{speed} = |\text{velocity}|$

3.

The Derivative

INTUITIVE DERIVATIVE
(NOTATION)
The derivative of $\,f\,$ at $\,x\,$ is the slope of the tangent line to the graph of $\,f\,$ at $\,(x,f(x))\,$; it is denoted by $$ \underbrace{f'(x)}_{\text{‘$\,f\,$ prime of $\,x\,$’}}\ \ \ (\text{prime notation})$$ or $$ \frac{d}{dx} f(x)\ \ \ (\text{Leibnitz notation})$$ How can a fct [function] fail to be differentiable?
a “kink”         a “break”         a vertical tangent line

EXAMPLE
[labels on graph] $\,x\,,$ $\,y\,,$ $\,y = f(x)\,$

$$\begin{alignat}{2} &f'(-\frac 12) = -10 && f'(2.1) \lt 0\\ &f'(0) \text{ DNE (does not exist)} &\quad& f'(3) = 0\\ &f'(\frac 12) = 10 && f'(3.3) > 0\\ &f'(1) \text{ DNE} && f'(4) \text{ DNE}\\ &f'(1.9) = 0 && f'(5) = 0 \end{alignat} $$

4.

Limits—The Idea
Introduction to Limits

LIMIT OF A FUNCTION
We write $$\lim_{x\rightarrow a} f(x) = \ell\ \text{(“ell”)}$$ and say
“the limit of $\,f(x)\,,$ as $\,x\,$ approaches $\,a\,,$ equals $\,\ell\,$”
iff [if and only if]
we can make the values of $\,f(x)\,$ as close to $\,\ell\,$ as we like, by taking $\,x\,$ to be sufficiently close to ‘$\,a\,$’
(but not equal to ‘$\,a\,$’)


3 CASES WHERE $\,\underbrace{\lim_{x\rightarrow a} f(x) = \ell}_{\text{This sentence is true!}}\,$
[labels on all three graphs] $\,a\,,$ $\,\ell\,,$ $\,y = f(x)\,$
(In all 3 cases, the fct [function] $\,f\,$ is defined near $\,a\,.$ )
Alternate notation: As $\,\underbrace{x\rightarrow a}_{\text{$\,x\,$ approaches $\,a\,$}}\,,$ $\,f(x)\rightarrow\ell\,.$

5.

Limits—Making it Precise

ONE-SIDED LIMITS
[iff = if and only if]
$$\lim_{x\rightarrow a^+} f(x) = \ell\ \ \text{iff}\ \ \left(\text{as } x\rightarrow a^+\,, \ f(x)\rightarrow \ell\right)$$ Read: The limit of $\,f(x)\,,$ as $\,x\,$ approaches $\,a\,$ from the right, equals $\,\ell\,$
(similar for $\,x\rightarrow a^-\,$)

Example:
$\displaystyle\lim_{x\rightarrow 2^-} f(x) = 3\,,$     $\,f(2) = 4\,$
$\displaystyle\lim_{x\rightarrow 2^+} f(x) = 5\,,$     $\displaystyle\lim_{x\rightarrow 2} f(x) \text{ DNE}\,$
[labels on graph] $\,2\,$ (on $x$-axis), $\,y = f(x)\,$

THE SQUEEZE (PINCHING) THEOREM
If $\,f(x)\le g(x)\le h(x)\,$ when $\,x\,$ is near $\,a\,$ (except possibly at $\,a\,$) and $$\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} h(x) = \ell$$ then   $\displaystyle\lim_{x\rightarrow a} g(x) = \ell\,$
[labels on graph] $\,a\,,$ $\,\ell\,,$ $\,f\,,$ $\,g\,,$ $\,h\,$
EX:   $\displaystyle \lim_{x\rightarrow 0} x^2\sin\frac 1x = 0\,$ since $$-x^2 \le x^2\sin\frac 1x \le x^2$$ [show $\,-x^2\,$ and $\,x^2\,$ both going to $\,0\,$]

6.

Continuity

CONTINUITY AT A POINT
[iff = if and only if]
$$f\text{ is continuous at } c\ \ \text{iff}\ \ \underbrace{\lim_{x\rightarrow c} f(x) = f(c)}_{\text{(*)}}$$ (*) says three things:

1) $\,f\,$ is defined at $\,c\,$ (i.e., $\,f(c)\,$ exists)
2) $\displaystyle\lim_{x\rightarrow c} f(x)\,$ exists (as $\,x\rightarrow c\,,$ $\,f(x)\,$ gets close to something)
3) (1) and (2) are equal

Roughly: a fct [function] is cont [continuous] at $\,c\,$ if there is no break in the graph at $\,c\,$

THREE WAYS A FCT [function] CAN FAIL TO BE CONTINUOUS AT $\,c\,$

(1) $\,f\,$ is not defined at $\,c\,$

(2) “nonremovable” discontinuities,
$\displaystyle\lim_{x\rightarrow c} f(x)\,$ does not exist

(3) both $\,f(c)\,$ and $\displaystyle \lim_{x\rightarrow c} f(x)\,$ exist, but are not not equal (a “removable” discontinuity)

7.

Properties of Limits

KEY IDEA
If $\,f\,$ is continuous at $\,c\,,$ then $$ \lim_{x\rightarrow c} f(x) = \underbrace{f(c)}_{\text{evaluate $\,f\,$ at $\,c\,$}} $$ That is, evaluating a limit
is as easy as DIRECT SUBSTITUTION!!
Note: We say $\,f\,$ is a continuous function if $\,f\,$ is continuous at every pt [point] in its domain.

EXAMPLE
$$ \begin{align} \lim_{x\rightarrow 1} (x^3 &- 2x^2 + 5x - 7)\cr &= 1^3 - 2\cdot 1^2 + 5\cdot 1 - 7\cr &= -3 \end{align} $$ Polynomials are continuous everywhere (no breaks), so evaluating limits is as easy as direct substitution!

8.

Instantaneous Rates of Change

AN IMPORTANT TYPE
OF CONTINUITY PROBLEM
Determine the values of $\,a\,$ and $\,b\,$ if $\,f\,$ is continuous: $$ f(x) = \begin{cases} x^2\,, &\ \ &x \lt -4\cr ax+b\,, &&-4\le x \lt 5\cr \sqrt{x+31}\,,&&5\le x \end{cases} $$ SOLN: [Solution]
$$\begin{gather} f(-4) = -4a + b \overset{\text{must}}{=} (-4)^2 = 16\cr f(5) = \sqrt{5 + 31} = 6 \overset{\text{must}}{=} 5a+b \end{gather} $$ Solve the system:

$$ \begin{align} -4a + b &= 16\cr 5a + b &= 6 \end{align} $$

to get...

$$ \begin{align} a = -10/9\cr b = 104/9 \end{align} $$

RELATIONSHIP BETWEEN DIFFERENTIABILITY AND CONTINUITY
A function can be continuous at a pt [point], but not differentiable there: [picture of kink]

However:

THEOREM. If $\,f\,$ is differentiable at $\,c\,,$
then $\,f\,$ is continuous at $\,c\,.$

So... “differentiability is STRONGER than continuity!”

9.

Properties of Limits

LIMIT RULES
Spse [suppose] that $\displaystyle\,\lim_{x\rightarrow a} f(x)\,$ and $\displaystyle\,\lim_{x\rightarrow a} g(x)\,$ both exist. Then: $$ \lim_{x\rightarrow a} \bigl(f(x) + g(x)\bigr) = \lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x) $$ (Note the different order of operations!)
“The limit of a sum is the sum of the limits,
providing both individual limits exist” $$ \begin{gather} \lim_{x\rightarrow a} f(x)g(x) = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x)\cr\cr \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow a} f(x)}{\lim_{x\rightarrow a} g(x)} \quad \bigl(\lim_{x\rightarrow a} g(x) \ne 0\,,\ \ g(x)\ne 0\text{ near } a\bigr) \end{gather} $$ Verbalize!

TECHNIQUES FOR EVALUATING LIMITS
(EXAMPLES)
[fct = function; CONT = continuous; @ = at] $$ \begin{gather} \lim_{x\rightarrow 0} \frac{-3-2x+x^2}{x+1} = \frac{-3}1 = -3 \ \ (\text{fct is cont @ $0$; direct substitution})\cr \lim_{x\rightarrow -1}\frac{-3-2x+x^2}{x+1} = \lim_{x\rightarrow -1} \frac{\cancel{(x+1)}(x-3)}{\cancel{x+1}} = -1-3 = -4 \end{gather} $$ [reln = relationship; poly'ls = polynomials; $\,\forall = \,$ for all;
iff = if and only if]
(Recall reln between zeroes and factors of poly'ls; $\,f(c) = 0\iff x-c\ \text{ is a factor of $\,f(x)\,$}\,$: and, $\,\frac{x+1}{x+1} = 1\ \ \forall \ \ x\ne -1\,$) $$ \lim_{x\rightarrow\infty} \frac{x^2-2x-1}{\sqrt{2x^4 - 2}}\cdot\frac{\frac{1}{x^2}}{\frac{1}{\sqrt{x^4}}} = \lim_{x\rightarrow\infty}\frac{1 - \frac{2}{x}- \frac{1}{x^2}}{\sqrt{2-\frac{2}{x^4}}} = \frac{1}{\sqrt 2} $$ “Divide by largest power of $\,x\,$ trick”
to handle $\,\frac{\infty}{\infty}\,$ “indeterminate form” $$ \lim_{x\rightarrow\infty} (\sqrt{x+1} - \sqrt{x})\cdot\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}} = \lim_{x\rightarrow\infty} \frac{x+1-x}{\sqrt{x+1}+\sqrt{x}} = 0 $$

10.

Indeterminate Forms

INDETERMINATE FORMS
“Cannot be determined without additional work!”
Limits of the form $\,\frac{0}{0}\,,$ $\,\frac{\pm\infty}{\pm\infty}\,,$ $\,(0)(\pm\infty)\,,$ $\,\infty - \infty\,,$ $\,0^0\,,$ $\,(\pm\infty)^0\,,$ and $\,1^\infty\,$ may exist, and may equal any real #, or may not exist!
[left of top card] (What #s are each of these “competing” between?)

	“$\frac{0}{0}$”	“$\frac{\pm\infty}{\pm\infty}$”
EXISTS	$\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$	$\displaystyle\lim_{x\rightarrow\infty}\frac{x^2-1}{2x^2+1} = \frac 12$
DNE or $\,\pm\infty\,$	$\displaystyle\lim_{x\rightarrow 0}\frac{x}{x^2}\,$ DNE	$\displaystyle\lim_{x\rightarrow\infty}\frac{x^2}x = \infty$

DECIDING IF YOU HAVE AN “INDETERMINATE FORM” (OR NOT!)
Is “$\,0^\infty\,$” an indeterminate form?
Consider $\,x^y\,$: Hold $\,x\,$ close to $\,0\,,$
let $\,y\rightarrow\infty\,,$
so $\,x^y\rightarrow 0\,$
( like $\,(\frac 1{100})^\infty$ )
Then, hold $\,y\,$ close to $\,\infty\,,$
let $\,x\rightarrow 0\,,$
so $\,x^y\rightarrow 0\,$
( like $\,(0)^{(10^{18})} )$
In both cases, the outputs $\,\rightarrow 0\,.$
They're not competing;
this is not an indeterminate form!
[right side] Example: $\,\frac00\,$
numerator $\,\rightarrow 0\,$ is trying to make outputs $\,\rightarrow 0\,$
denominator $\,\rightarrow 0\,$ is trying to make outputs big!!

11.

The Derivative

DEFINITION:
THE DERIVATIVE OF $\,f\,$ AT $\,a\,$
$$ \begin{align} \underbrace{f'(a)}_{\text{$f\,$ prime of $\,a$}} &= \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}\cr &= \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a} \end{align} $$ [top graph labels: points $\,(a,f(a))\,$ and $\,(a+h,f(a+h))\,,$ $\,h\,$]
[bottom graph labels: points $\,(a,f(a))\,$ and $\,(x,f(x))\,$]
$f'(a)\,$ is the “derivative of $\,f\,,$ at $\,a\,$”; when $\,f'(a)\,$ exists, we say that “$\,f\,$ is differentiable at $\,a\,$”.

USING THE DEFINITION OF DERIVATIVE
Let $\,f(x) = x^2-2x+1\,.$ Find $\,f'(3)\,$ and $\,f'(x)\,,$
using the definition of derivative. $$\begin{align} f'(3) &= \lim_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h} = \lim_{h\rightarrow 0}\frac{(3+h)^2-2(3+h)+1 - (3^2-2\cdot3 + 1)}{h}\cr &= \lim_{h\rightarrow 0}\frac{\cancel{9} + 6h + h^2 -\cancel{6} - 2h + \cancel{1} - \cancel{4}}{h} = \lim_{h\rightarrow 0}\frac{\cancel{h}(4+h)}{\cancel{h}} = 4\cr\cr f'(x) &= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{(x+h)^2-2(x+h)+1 - (x^2-2x + 1)}{h}\cr &= \lim_{h\rightarrow 0}\frac{\cancel{x^2} + 2xh + h^2 -\cancel{2x} - 2h + \cancel{1} - \cancel{x^2} + \cancel{2x} - \cancel{1}}{h}\cr &= \lim_{h\rightarrow 0}\frac{\cancel{h}(2x +h - 2)}{\cancel{h}} = 2x - 2 \end{align} $$ Not really any more work to get the formula
for all values of $\,x\,$!!

12.

The Derivative

AN ALTERNATE APPROACH
(YOUR CHOICE! COMPARE WITH 11b)
$$ \begin{align} f'(x) &= \lim_{y\rightarrow x} \frac{f(y)-f(x)}{y-x} = \lim_{y\rightarrow x}\frac{y^2-2y+1 - (x^2-2x+1)}{y-x}\cr &= \lim_{y\rightarrow x}\frac{y^2 - x^2 -2y+2x}{y-x} = \lim_{y\rightarrow x}\frac{(y-x)(y+x)-2(y-x)}{y-x}\cr &= \lim_{y\rightarrow x}\frac{\cancel{(y-x)}(y+x-2)}{\cancel{(y-x)}} = x + x - 2 = 2x - 2 \end{align} $$
ANOTHER EXAMPLE:
USING THE DEFN OF DERIVATIVE
Let $\,f(x) = \frac 1x\,.$
$$ \begin{align} f'(x) &= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}h = \lim_{h\rightarrow 0}\frac{\frac{1}{x+h} - \frac 1x}{h}\cr &= \lim_{h\rightarrow 0}\frac{\frac{1}{x+h}\cdot\frac xx - \frac 1x\frac{(x+h)}{(x+h)}}{h}\cr &= \lim_{h\rightarrow 0}\frac{x-(x+h)}{(x+h)xh} = \lim_{h\rightarrow 0}\frac{-\cancel{h}}{\cancel{h}x(x+h)}\cr &= -\frac{1}{x^2} \end{align} $$

13.

Tangent Lines

THE LINEARIZATION OF $\,f\,$ AT $\,a\,$
... is the best linear approximation to $\,f\,$ at the point $\,(a,f(a))\,$; it's the tangent line approximation.
Equation (use pt-slope fm [point-slope form]): $$y - f(a) = f'(a)(x-a)$$ Solve for $\,y\,$; re-name: $$\ell_a(x) = f(a) + f'(a)(x-a) $$ [sketch at left: $\,y = \ell_a(x)\,$; point $\,(a,f(a))\,$;
$m = f'(a)\,$; $\,y = f(x)\,$]
[circles at bottom: zoom, zoom, looks linear!!]

INTUITION FOR THE LINEARIZATION
$$ \underbrace{\ell_a(x)}_{\substack{\text{nearby fct values}\\ \text{are best approximated by...}}} = \underbrace{f(a)}_{\text{the value @ ‘$\,a\,$’}} + \underbrace{ \overbrace{\underbrace{f'(a)}_{\substack{\text{rate of}\\ \text{change}\\ \text{at ‘$\,a\,$’}} }}^{\substack{\text{how fast the fct values}\\ \text{are changing at ‘$\,a\,$’}}} \cdot \underbrace{(x-a)}_{\text{change in $\,x$}} }_{\substack{\text{how much $\,f\,$ has changed by,}\\ \text{as the input changes from ‘$\,a\,$’ to ‘$\,x\,$’}}} $$ Note: For $\,x\,$ near $\,a\,,$ $\,f(x)\cong \ell_a(x)\,$

14.

Local Max/Min
Graphing Functions

LOCAL MAX/MIN (EXTREME VALUES)
$$ \underbrace{\text{MAXIMUM/MINIMUM}}_{\text{SINGULAR}} \quad \underbrace{\text{MAXIMA/MINIMA}}_{\text{PLURAL}} $$ [sketches: point $\,(c,f(c))\,$]
[iff = if and only if]
A fct [function] $\,f\,$ has a local max @ [at] $\,c\,$ iff
$\,\exists\,$ [there exists] an open interval $\,I\,$ containing $\,c\,$
s.t. [such that] $\,\forall\ \ x\in I,\ \ f(x)\le f(c)\,$;
the # $\,f(c)\,$ is the local max;
it occurs at $\,x = c\,$
[side brace]: nearby fct values are lower (or the same)

A fct $\,f\,$ has a local min @ $\,c\,$ iff
$\,\exists\,$ an open interval $\,I\,$ containing $\,c\,$
s.t. $\,\forall\ \ x\in I,\ \ f(x)\ge f(c)\,$
[side brace]: nearby fct values are higher (or the same)

GLOBAL MAX/MIN
$\,f\,$ has a global max at $\,a\,$ iff $\,f(a) \ge f(x)\ \ \forall\ \ x\in\text{dom}(f)\,$;
$\,f\,$ has a global min at $\,a\,$ iff $\,f(a) \le f(x)\ \ \forall\ \ x\in\text{dom}(f)\,$;

[left sketch labels]: $\,a\,,$ $\,b\,,$ $\,2\,,$ $\,3\,,$ $\,5\,$
[beneath left sketch]:
$\,2\,$ is a local min; at $\,x=a\,$
$\,2\,$ is a global min; at $\,x=a\,$
$\,3\,$ is a local max; at $\,x = b\,$
No global max

[right sketch labels]: $\,a\,,$ $\,b\,,$ $\,c\,,$ $\,d\,,$ $\,e\,,$ $\,4\,,$ $\,6\,,$ $\,7\,,$ $\,10\,$
[beneath right sketch]:
$\,10\,$ is a local and global max; @ $\,x = b\,$ and $\,x = d\,$
$\,7\,$ is a local min, @ $\,x = c\,$
$\,6\,$ is a local min, @ $\,x = e\,$
$\,4\,$ is a local and global min, @ $\,x = a\,$

15.

Increasing/Decreasing Functions

STRICTLY INCREASING FCTS [functions] /
STRICTLY DECREASING FCTS
[iff = if and only if]
A fct $\,f\,$ is strictly increasing on an interval $\,I\,$ iff whenever $\,x \lt y\,$ in $\,I\,,$ $\,f(x) \lt f(y)\,$
[sketch at left]: points $\,(x,f(x))\,$ and $\,(y,f(y))\,,$ (“uphill” graphs)

A fct $\,f\,$ is strictly decreasing on an interval $\,I\,$ iff whenever $\,x \lt y\,$ in $\,I\,,$ $\,f(x) > f(y)\,$
[sketch at left]: points $\,(x,f(x))\,$ and $\,(y,f(y))\,,$ (“downhill” graphs)

GETTING INC/DEC [increasing/decreasing]
BEHAVIOR FROM THE DERIVATIVE
Spse [suppose] $\,f\,$ is differentiable on an interval $\,I\,.$
If $\,f'(x) > 0\,\ \overbrace{\text{ on $\,I\,$}}^{\forall\ \ x\in I}\,,$ then $\,f\,$ is strictly increasing on $\,I\,$

If $\,f'(x) \lt 0\,$ on $\,I\,,$ then $\,f\,$ is strictly decreasing on $\,I\,$
[on side]: $\,\forall = \,$ “For all”

16.

the Second Derivative

CONCAVITY
[iff = if and only if]
[sketches]: “holding water” concave up;
“shedding water” concave down

Definitions:
$\,f\,$ is concave up on $\,I\,$
iff $\,f'\,$ increases on $\,I\,$
(slopes increase as you move from left to right)

$\,f\,$ is concave down on $\,I\,$
iff $\,f'\,$ decreases on $\,I\,$
(slopes decrease as you move from left to right)

GETTING CONCAVITY INFO FROM
THE SECOND DERIVATIVE
$\,f = \,$ function,   $\,f' = \,$ 1st deriv [derivative],
$\,f'' = \,$ 2nd deriv
$$ f \text{ is concave up on } I \ \ \ \ \text{iff}\ \ \ \ f''(x) > 0 \text{ on } I $$ (Spse [suppose] $\,f\,$ is twice differentiable on $\,I\,.$ )
$$ f \text{ is concave down on } I \ \ \ \ \text{iff}\ \ \ \ f''(x) \lt 0 \text{ on } I $$

17.

Basic Differentiation Formulas

POWER RULE FOR DIFFERENTIATION

If $\,n\,$ is any real number, then $$ \frac d{dx} (x^n) = nx^{n-1} $$

Note: The “$\,\frac{d}{dx}\,$” operator;
“take the derivative, with respect to $\,x\,$”

Believable? Look at 2 special cases:
$y = x^3\,,$ right shape for $\,3x^2\,$
$\,y = x^4\,,$ right shape for $\,4x^3\,$

EXAMPLES: USING THE POWER RULE

• $\frac{d}{dx}(x^{1.7}) = 1.7x^{1.7-1} = 1.7x^{0.7}$
• If $\,f(x) = x^{-4}\,,$ then $\,f'(x) = (-4)x^{-4-1} = -4x^{-5}$
• $\frac{d}{dt} t^{\sqrt 2} = \sqrt{2} t^{\sqrt 2 - 1}\,$
  (use $\,\frac{d}{dt}\,$ with functions of $\,t\,$)
• Let $f(x) = \root 3\of{x}\,$; find $\,f'(x)\,$:
  $f(x) = \root 3\of{x} = x^{1/3}\,$;
  $f'(x) = \frac 13x^{\frac 13 - 1} = \frac 13x^{-2/3} = \frac{1}{3x^{2/3}} = \frac{1}{3\root 3\of{x^2}}$

Try to write the derivative in a form
that matches the original function.

18.

Basic Differentiation Formulas

PARTIAL PROOF OF THE POWER RULE
Recall the “telescopic identities”:
$$ \begin{align} a^2 - b^2 &= (a-b)(a+b)\cr a^3 - b^3 &= (a-b)(a^2 + ab + b^2)\cr a^4 - b^4 &= (a-b)(a^3 + a^2b + ab^2 + b^3)\cr &\vdots\cr a^n - b^n &= (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}) \end{align} $$ Note: $P(a) = a^n - b^n\,$
$b\,$ is a zero, so $\,a - b\,$ is a factor $$ \begin{align} &a^2 + ab + b^2\cr a - b \ \ \Rule{1px}{15px}{1px}\ &\overline{a^3 - b^3}\cr &\kern-11pt\underline{-(a^3 - ba^2)}\cr &ba^2 - b^3\cr &\kern-11pt\underline{-(ba^2 - ab^2)}\cr &ab^2 - b^3\cr &\kern-11pt\underline{-(ab^2 - b^3)}\cr &0 \end{align} $$ (Note the pattern that occurs in long division)

(side b)
Let $\,n = 0,1,2,...\,$

Clearly, $\,\frac{d}{dx} x^0 = 0x^{0-1} = 0\,,$
since $\,x^0\,$ is a constant function.

Also, $\,\frac{d}{dx} x^1 = 1x^{1-1} = 1x^0 = 1\,,$
since $\,x\,$ is a line with slope $\,= 1\,.$

Let $\,f(x) = x^n\,$ for $\,n \ge 2\,$; then $$ \begin{align} f'(x) &= \lim_{y\rightarrow x} \frac{y^n - x^n}{y-x}\cr &= \lim_{y\rightarrow x}\frac{(y-x)(y^{n-1} + y^{n-2}x + \cdots + x^{n-1})}{y-x}\cr &= \lim_{y\rightarrow x}\ (y^{n-1} + y^{n-2}x + \cdots + yx^{n-2} + x^{n-1})\cr &= x^{n-1} + \underbrace{x^{n-2}x^1 + \cdots + x^1x^{n-2} + x^{n-1}}_{n-1 \text{ terms}}\cr &= nx^{n-1} \end{align} $$

19.

Basic Differentiation Formulas

DERIVATIVE OF A CONSTANT
TIMES A FCT [function] $$\frac d{dx} cf(x) = c\cdot\frac d{dx} f(x)$$ [arrow showing $\,c\,$ being ‘slid out’]
Different orders of operation on each side.
You can “slide constants out”.

Pf. [proof] Spse [suppose] $\,f\,$ is differentiable at $\,x\,,$ and let $\,g(x) = cf(x)\,.$ Then, $$ \begin{align} g'(x) &= \lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{h}\cr &= \lim_{h\rightarrow 0} \frac{cf(x+h) - cf(x)}{h}\cr &= \lim_{h\rightarrow 0}c\cdot\frac{f(x+h)-f(x)}{h}\cr &= c\cdot \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\ \ \text{(via properties of limits)}\cr &= c f'(x) \end{align} $$ FINDING $\,\frac{d}{dx}(Kx^n)\,$
$$ \begin{alignat}{2} \frac d{dx} Kx^n &= K\frac d{dx} x^n &\ \ &\text{(slide constant out)}\cr &= K(nx^{n-1}) &&\text{(power rule)}\cr \end{alignat} $$ so $$\frac d{dx} Kx^n = nKx^{n-1}$$ EX [example]: $\,\frac d{dx} 3x^7 = 21x^6\,$
EX: If $\,A(r) = \pi r^2\,,$ then $\,A'(r) = 2\pi r\,$

20.

Basic Differentiation Formulas

DERIVATIVES OF SUMS AND DIFFERENCES
$$ \underbrace{\frac d{dx} \bigl( f(x) + g(x)\bigr)}_{\substack{\text{the derivative}\\ \text{of a sum}}} \ \ \underbrace{\strut =}_{\text{ is }} \ \ \overset{\swarrow\ \text{the derivatives}\ \searrow} { \frac d{dx} f(x) \underbrace{\strut +}_{\text{the sum of}} \frac d{dx} g(x) } $$

Then:

$$ \begin{align} \frac d{dx} &\bigl( f(x) - g(x) \bigr)\cr &= \frac d{dx} \bigl( f(x) + (-1)g(x)\bigr)\ \ \text{(subtraction is a special kind of addition)}\cr &= \frac d{dx} f(x) + \frac d{dx} (-1)g(x)\ \ \text{[arrow showing $\,-1\,$ being slid out]}\cr &= \frac d{dx} f(x) + (-1)\frac d{dx} g(x)\cr &= \frac d{dx} f(x) - \frac d{dx} g(x) \end{align} $$ PROOF
Let $\,S(x) = f(x) + g(x)\,$; spse [suppose] $\,f\,$ and $\,g\,$ are both differentiable at $\,x\,.$ Then, $$ \begin{align} S'(x) &= \lim_{h\rightarrow 0} \frac{S(x+h)-S(x)}{h}\cr &= \lim_{h\rightarrow 0} \frac{f(x+h) + g(x+h) - (f(x) + g(x))}{h}\cr &= \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} + \frac {g(x+h)-g(x)}{h}\cr &= \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}h + \lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}h\ \ \text{(property of limits)}\cr &= f'(x) + g'(x)\ \ \text{(defn [definition] of derivative)} \end{align} $$

21.

Differentiating Composite Functions

DIFFERENTIATING COMPOSITE FUNCTIONS:
MOTIVATION
Ray runs 2 times as fast as Carol,
and Carol runs 3 times as fast as Julia,
so Ray runs 6 times as fast as Julia.
Rates multiply!

DIFFERENTIATING COMPOSITE FUNCTIONS:
THE IDEA
[labels on sketch:
$\,x\,$ goes into $\,g\,$; Rate of change: $\,g'(x)\,$
$\,g(x)\,$ goes into $\,f\,$: Rate of change: $\,f'(g(x))\,$
big box labeled $\,f\circ g\,$; final output $\,f(g(x))\,$]
So how fast is $\,f(g(x))\,$ changing with respect to $\,x\,$? Multiply these!! [$\,g'(x)\,$ and $\,f'(g(x))\,$]

22.

Differentiating Composite Functions

THE CHAIN RULE
(PRIME NOTATION)
Recall: $\,(f\circ g)(x) = f(g(x))\,$

Then:

$$ \begin{align} (f\circ g)'(x) &= f'(g(x))\cdot g'(x)\cr \text{or }\ \frac d{dx} f(g(x)) &= \underbrace{f'(g(x))}_{\substack{\text{how fast $\,f\,$ is changing}\\ \text{wrt [with respect to] $\,g\,$}}} \underset{\text{roughly:}}{\cdot } \underbrace{g'(x)}_{\substack{\text{how fast $\,g\,$ is changing}\\ \text{wrt $\,x\,$}}} \end{align} $$ THE CHAIN RULE
(LEIBNITZ NOTATION)
Spse [suppose] $\,y = f(u)\,$ and $\,u = g(x)\,$
(   so $\,y = f(g(x))\,$   )

Then,

$$ \frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} $$ [dashed line ‘cancelling out’ out the ‘$\,du\,$’ elements:] Memory device; NOT cancellation!

23.

Differentiating Composite Functions

WHY IS IT CALLED THE “CHAIN RULE”?
$$ \begin{align} \frac d{dx} a(&b(c(d(x))))\cr &= \underbrace{a'(b(c(d(x))))\cdot b'(c(d(x)))\cdot c'(d(x))\cdot d'(x)}_{\text{a long chain!}} \end{align} $$ $$ \begin{gather} y = f_1(t)\,,\ \ \ t = f_2(u)\,,\ \ \ u = f_3(v)\,,\ \ \ v = f_4(w)\cr \frac{dy}{dw} = \frac{dy}{dt}\cdot\frac{dt}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dw} \end{gather} $$
USING THE CHAIN RULE
TO DIFFERENTIATE $\,(f(x))^n\,$

Let $\,\underbrace{g(x) = x^n}_{g'(x) = nx^{n-1}}\,,$ so $\,g(f(x)) = (f(x))^n\,.$ Then, $$ \begin{align} \frac d{dx} (f(x))^n &= \frac d{dx} g(f(x))\cr &= g'(f(x))\cdot f'(x)\ \ \text{(chain rule)}\cr &= n(f(x))^{n-1}\cdot f'(x)\cr \end{align} $$ So $$\boxed{\frac d{dx} (f(x))^n = n(f(x))^{n-1}\cdot f'(x)}$$

24.

Differentiating Composite Functions

PATTERN FOR GENERALIZING ALL THE
BASIC DIFFERENTIATION FORMULAS
... replace $\,x\,$ by $\,f(x)\,,$ and multiply by $\,f'(x)\,$!!

Basic Formula	Generalization
$$\frac d{dx} x^n = nx^{n-1}$$	$$\frac d{dx} (f(x))^n = n(f(x))^{n-1}\cdot f'(x)$$
$$\frac d{dx} {\text{e}}^x = {\text{e}}^x$$	$$\frac d{dx} {\text{e}}^{f(x)} = {\text{e}}^{f(x)}\cdot f'(x)$$
$$\frac d{dx} \sin x = \cos x$$	$$\frac d{dx} \sin f(x) = (\cos f(x))\cdot f'(x)$$

EXAMPLES

1. $\displaystyle\frac d{dx} (3x+1)^7 = 7(3x+1)^6\cdot 3 = 21(3x+1)^6$
2. Let $\displaystyle\,f(x) = \frac 1{\sqrt{2x-3}} = (2x-3)^{-1/2}\,$;
   then $\displaystyle\,f'(x) = -\frac 12(2x-3)^{-3/2}\cdot 2 = \frac{-1}{\sqrt{(2x-3)^3}}$
   (Try to write the derivative in a form that matches the original function.)
3. 
   $$ \begin{align} \frac d{dx} (5x^7 - \frac{4}{x+3}) &= 35x^6 - 4\frac d{dx} (x+3)^{-1}\cr &= 35x^6 - 4(-(x+3)^{-2})\cr &= 35x^6 + \frac 4{(x+3)^2} \end{align} $$

25.

Differentiating Products/Quotients

the PRODUCT RULE FOR DIFFERENTIATION
If $\,f\,$ and $\,g\,$ are both differentiable at $\,x\,,$ then $$ \underbrace{\frac{d}{dx}\bigl( f(x)g(x) \bigr)}_{\text{in words: }\ \substack{\text{the derivative}\cr \text{of a product}}} \ \underbrace{=\strut}_{\text{is}}\ \underbrace{f(x)\strut}_{\text{the first}}\ \underbrace{g'(x)}_{\substack{\text{times the}\cr\text{derivative}\cr\text{of the}\cr\text{second}}} \ \underbrace{+\strut}_{\text{plus}} \ \underbrace{g(x)}_{\text{the second}}\ \underbrace{f'(x)}_{\substack{\text{times the}\cr\text{derivative}\cr\text{of the}\cr\text{first}}} $$ Caution: The derivative of a product is NOT the product of the derivatives!!

PF [proof] OF THE PRODUCT RULE
Let $\,f\,$ and $\,g\,$ be differentiable at $\,x\,,$ and let $\,P(x) = f(x)g(x)\,.$ Then, $$ \begin{align} P(x) &= \lim_{h\rightarrow 0} \frac{P(x+h)-P(x)}h\cr &= \lim_{h\rightarrow 0} \frac{f(x+h)g(x+h) \ \overbrace{- f(x+h)g(x) + f(x+h)g(x)}^{\text{add zero!}} - f(x)g(x)}h\cr &\overset{\text{algebra}}{=} \lim_{h\rightarrow 0}f(x+h)\frac{g(x+h)-g(x)}h + \frac{f(x+h)-f(x)}h g(x)\cr &= \lim_{h\rightarrow 0} f(x+h) \lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}h + \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}h\lim_{h\rightarrow 0} g(x)\cr &= \strut f(x)g'(x) + f'(x)g(x) \end{align} $$

26.

Continuity

WHEN IS $\displaystyle\,\lim_{h\rightarrow 0} f(x+h) = f(x)\,$?
Answer: Not always!

[sketch: hollow dot $\,(x,1)\,,$ point $\,(x,2)\,,$
point on curve $\,(x+h,f(x+h))\,,$
curve labeled $\,y = f(x)\,$]

As $\,h\rightarrow 0\,,$ $\,f(x+h)\rightarrow 1\,$
but $\,f(x) = 2\,$
so $\displaystyle \underbrace{\lim_{h\rightarrow 0} f(x+h)}_{1} \ne \underbrace{\strut f(x)}_{2}\,$ $$ \lim_{h\rightarrow 0} f(x+h) = f(x)\ \ \text{iff}\ \ \text{$\,f\,$ is continuous at $\,x\,$} $$
WHEN IS THIS TRUE?
As $\,x\rightarrow c\,,$ $\,f(x)\rightarrow f(c)\,$
Answer: Not always!

[sketch: hollow dot $\,(c,1)\,,$ point $\,(c,2)\,,$ $\,x\,,$ $\,y = f(x)\,$]

As $\,x\rightarrow c\,,$ $\,f(x)\rightarrow 1\,$; but $\,f(c) = 2\,.$ $$ \text{As } x\rightarrow c\,, f(x)\rightarrow f(c)\quad\text{iff}\quad \text{$\,f\,$ is continuous at $\,c\,$} $$

27.

Differentiating Products/Quotients

THE QUOTIENT RULE
FOR DIFFERENTIATION
Spse [suppose] $\,f\,$ and $\,g\,$ are differentiable at $\,x\,,$
and $\,g(x)\ne 0\,.$ Then, $$ \frac d{dx}\biggl(\frac{f(x)}{g(x)}\biggr) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} $$ “The derivative of a quotient is:
the bottom, times the derivative of the top,
minus the top, times the derivative of the bottom,
all over the bottom squared” $$ \frac{d}{dx}\bigl(\frac{\text{HI}}{\text{HO}}\bigr) = \frac{\text{HO}\,d\text{HI} - \text{HI}\,d\text{HO}}{\text{HO HO}} $$ (it it helps!!)

PF [proof] OF THE QUOTIENT RULE
Let $\,Q(x) = \frac{f(x)}{g(x)} = f(x)\cdot [g(x)]^{-1}\,.$

Then,

$$ \begin{align} Q'(x) &= f(x)\cdot (-1)[g(x)]^{-2}g'(x) + [g(x)]^{-1}\cdot f'(x)\cr &= \frac{-f(x)g'(x)}{(g(x))^2} + \frac{f'(x)}{g(x)}\cdot\frac{g(x)}{g(x)}\cr &= \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \end{align} $$

28.

Indeterminate Forms

THE IRRATIONAL NUMBER $\,\text{e}\,$
What happens to $\,(1 + \frac 1n)^n\,$ as $\,n\rightarrow\infty\,$?
You might think it gets close to $\,1\,,$ since $\,\frac 1n\rightarrow 0\,$ and $\,1^{\text{any power}} = 1\,.$ But, you can't let the “inside” $\,n\,$ get big first, and then let the “outside” $\,n\,$ get big...

$\,n\,$	$\,(1 + \frac 1n)^n$	As $\,n\rightarrow\infty\,,$ $$(1+\frac 1n)^n \rightarrow \underbrace{\text{e}}_{\cong 2.72}$$
100	2.7048138
1000	2.7169239
10,000	2.7181459

EQUIVALENT LIMIT STATEMENTS
DEFINING $\,\text{e}\,$
[top sketch: vertical dashed line at $\,-1\,$;
horizontal dashed line at $\,\text{e}\,$; $\,y = (1+\frac 1x)^x\,$]
$$ \begin{gather} \lim_{x\rightarrow \infty} (1+\frac 1x)^x = \text{e}\cr \lim_{x\rightarrow -\infty} (1+\frac 1x)^x = \text{e}\cr \end{gather} $$ [bottom sketch: vertical dashed line at $\,-1\,$;
horizontal dashed line at $\,1\,$; hollow dot $\,(0,\text{e})\,$; $\,y = (1+x)^{1/x}\,$]
$$ \underbrace{\lim_{x\rightarrow 0}}_{\substack{\text{so both 1-sided limits}\cr\text{also exist, and equal $\,\text{e}\,$}}} \kern -0.4in(1+x)^{1/x} = \text{e} $$

29.

(included for
higher-level course)

EVALUATING A LIMIT INVOLVING $\,\text{e}\,$
FIND $\displaystyle\lim_{x\rightarrow 0^+} (1+kx)^{2/x}\,.$
SOLN [solution]: Let $\,kx := \frac 1n\,,$ so $\,n = \frac 1{kx}\,.$
As $\,x\rightarrow 0^+\,,$ $n\rightarrow\infty\,.$
Also, $\,\frac 1x = nk\,$ and so $\,\frac 2x = 2nk\,.$

Then,

$$ \lim_{x\rightarrow 0^+} (1+kx)^{2/x} = \lim_{n\rightarrow\infty} (1 + \frac 1n)^{2kn} = \lim_{n\rightarrow\infty} \left[(1+\frac 1n)^n\right]^{2k} = {\text{e}}^{2k} $$ WHY IS $\displaystyle \lim_{n\rightarrow\infty} \left[(1+\frac 1n)^n\right]^{2k} = {\text{e}}^{2k}\,$?
Let $\,f(x) = x^{2k}\,$; recall that $$\lim_{x\rightarrow\text{e}} f(x) = f(\text{e})\ \ \text{iff}\ \ \text{$\,f\,$ is continuous at $\,\text{e}\,$}\ . $$ The fct [function] $\,f\,$ is continuous at $\,\text{e}\,.$
And, as $\,n\rightarrow\infty\,,$ $\,(1+\frac 1n)\rightarrow\text{e}\,.$
So... continuity is at work here!!

30.

Basic Differentiation Formulas

DERIVATIVES OF LOGARITHMS
$$ \frac d{dx} (\log_a x) = \frac 1{x\ln a} $$ (allowable bases for logs: $\,a > 0\,,$ $\,a\ne 1\,$)

Special case: $$\frac d{dx}(\ln x) = \frac 1x$$ [sketch: at $\,x\,,$ tangent line has slope $\,m = \frac 1x\,$]
The slope of the tangent line
is the reciprocal of the $x$-value!

Generalize: $$ \frac d{dx} \ln f(x) = \frac 1{f(x)}\cdot f'(x) $$ Recall: Logarithms are only defined for positive $\,x\,$

PROOF THAT $\,\displaystyle\frac d{dx} (\ln x) = \frac 1x\,$
$$ \begin{alignat}{2} \frac d{dx} \ln x &= \lim_{h\rightarrow 0}\frac{\ln(x+h)-\ln x}h\cr &= \lim_{h\rightarrow 0}\frac{\ln(\frac{x+h}h)}h &\quad&\text{(property of logs)}\cr &= \lim_{h\rightarrow 0}\frac 1h\ln(1+\frac hx)\cr &= \lim_{y\rightarrow 0} \frac 1{xy}\ln(1+y)&&\text{let $\,y := \frac hx\,,$ so $\,\frac 1h = \frac 1{xy}\,$}\cr &= \lim_{y\rightarrow 0}\frac 1x\cdot \frac 1y\ln(1+y)&&\text{as $\,h\rightarrow 0\,,$ $\,y\rightarrow 0$}\cr &= \frac 1x \lim_{y\rightarrow 0}\ln \underbrace{(1+y)^{1/y}}_{\rightarrow\text{e}}\cr &\overset{\text{Why?}}{=} \frac 1x\ln(\text{e}) = \frac 1x \end{alignat} $$

31.

Basic Differentiation Formulas

MORE NOTATION FOR DERIVATIVES
If $\,y = f(t)\,,$ $$ \begin{align} y' &= \frac{dy}{dt}\cr y'' &= \frac d{dt}(\frac{dy}{dt}) = \frac{d^2y}{dt^2}\cr y''' &= \frac{d^3y}{dt^3}\cr y^{(4)} &= \frac{d^4y}{dt^4} \end{align} $$ [$y''\,$ and higher] Higher order derivatives
in Leibnitz notation
Usually switch notation at about the 4^th derivative

NOTATION: EVALUATING A DERIVATIVE
AT A POINT
$$ y'(3) = \frac{dy}{dt}\bigg|_{t=3} = \frac{dy}{dt}(3) $$ Read   ‘ $\,\bigg|_{t=3}\,$ ’   as:   “evaluated at $\,t = 3\,$” $$ \begin{align} \underbrace{\ln}_{\substack{\ln\ \text{is the}\cr\text{function}\cr\text{name}}}'(3) &= \frac d{dx}\ln(x)\bigg|_{x=3}\cr &= \frac d{dx}\ln(x)\bigg|_3 \quad \substack{ \text{this shorthand is}\cr \text{OK, if there's no}\cr \text{danger of confusion}} \end{align} $$

32.

Basic Differentiation Formulas

DIFFERENTIATING $\,\text{e}^x\,$
$$\,\boxed{\ \frac{d}{dx}{\text{e}}^x = {\text{e}}^x\ }\,$$ [sketch: point $\,(x,{\text{e}}^x)\,$ with tangent line]
The $y$-value of the pt [point] also tells the slope of the tangent line!

PF [proof]: For all $\,x\,,$ $$ \ln {\text{e}}^x = x\ln \text{e} = x $$ So, $\,\frac d{dx} \ln\text{e}^x = \frac d{dx} x\,$; that is, $$ \frac 1{\text{e}^x}\cdot \frac d{dx}\text{e}^x = 1 $$ Solve for $\,\frac d{dx}\text{e}^x\,$!!

DIFFERENTIATING $\,a^x\,$
$$ \boxed{\ \frac d{dx} a^x = a^x\ln a\ } $$ [sketch: $\,x\,$ into $\ \ln\ $ box, output $\,\ln x\,$;
$\,\ln x\ $ into $\,\text{e}^x\,$ box, output $\,{\text{e}}^{\ln x} = x\,$;
the boxes $\,\ln\,$ and $\,\text{e}^x\,$ are inverse fcts [functions]]

Pf. [proof] Use the identity $\,x = {\text{e}}^{\ln x}\,$ $$ \begin{align} \frac d{dx} a^x &= \frac d{dx} {\text{e}}^{\ln a^x} = \frac d{dx} {\text{e}}^{x\ln a}\cr\cr &= {\text{e}}^{x\ln a}\cdot \ln a = a^x\ln a \end{align} $$

33.

(included for
higher-level course)

WHAT IS $\,\displaystyle\underbrace{\lim_{x\rightarrow 0}\frac{\sin x}{x}}_{\text{“indeterminate form”}\ \frac 00}\,$?
[sketch: $\,y = \sin x\,,$ $\,y = x\,$]
MOTIVATION: The tangent line to the graph of $\,y = \sin x\,$ at $\,x = 0\,$ is $\,y = x\,$; so, for $\,x\,$ close to $\,0\,,$ $$\sin x\cong x\,;$$ it can be proven that $$\boxed{\ \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\ } $$ (When $\,x\,$ is small, can replace $\,\sin x\,$ with $\,x\,$
without much error!)
[side of card] Needed to prove that $\,\frac d{dx}\sin x = \cos x\,$

EXAMPLE
$$ \begin{align} \lim_{x\rightarrow 0} \frac{\sin(3x)}{x} &=\lim_{x\rightarrow 0} \frac{3\sin(3x)}{3x}\cr\cr &= 3\lim_{h\rightarrow 0}\frac{\sin h}h\quad\text{(let $\,h = 3x\,$; as $\,x\rightarrow 0\,,$ $\,h\rightarrow 0\,$)}\cr\cr &= 3 \end{align} $$

34.

(included for
higher-level course)

DERIVATIVES OF SINE AND COSINE
$$ \begin{gather} \boxed{\ \frac d{dx}\sin x = \cos x\ }\cr \boxed{\ \frac d{dx}\cos x = -\sin x\ } \end{gather} $$ [sketches: top graph labeled $\,y = \sin x\,$
Take derivative; right shape for cosine curve!
Take derivative again...]

Note: $\,\sin^2x = (\sin x)^2\,$
$\cos^4 x = (\cos x)^4\,,$ etc.

DERIVATIVES OF
TANGENT AND COTANGENT
$$ \begin{align} \frac d{dx}\tan x &= \frac d{dx}\left(\frac{\sin x}{\cos x}\right)\cr\cr &= \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x}\cr\cr &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \bigl(\frac 1{\cos x}\bigr)^2\cr\cr &= \sec^2 x \end{align} $$ so $$ \boxed{\ \frac d{dx}\tan x = \sec x \ } $$ and (you prove this!) $$ \boxed{\ \frac d{dx}\cot x = -\csc^2 x \ } $$ [side sketch: note the pairs that are reciprocals] $$ \begin{gather} \sin\cr \cos\cr \tan\cr \cot\cr \sec\cr \csc \end{gather} $$

35.

(included for
higher-level course)

DERIVATIVES OF
SECANT AND COSECANT
$$ \begin{align} \frac d{dx}\sec x &= \frac d{dx} \left(\frac 1{\cos x}\right)\cr\cr &= \frac{(\cos x)(0) - (1)(-\sin x)}{(\cos x)^2}\cr\cr &= \frac{\sin x}{\cos x}\cdot\frac 1{\cos x}\cr\cr &= \tan x\sec x \end{align} $$ so $$ \boxed{\ \frac d{dx}\sec x = \sec x\,\tan x\ } $$ also (you prove this!) $$ \boxed{\ \frac d{dx}\csc x = -\csc x\,\cot x\ } $$
A USEFUL OBSERVATION—
DERIVATIVES OF CO-FUNCTIONS
$$ \frac d{dx}\sin x = \cos x \qquad \frac d{dx}\cos x = -\sin x $$ [notes: replace by co-fct, add minus sign] $$ \frac d{dx}\tan x = \sec^2 x \qquad \frac d{dx}\cot x = -\csc^2 x $$ $$ \frac d{dx}\sec x = \sec x\,\tan x \qquad \frac d{dx}\csc x = -\csc x\,\cot x $$ Replace fcts [functions] by co-fcts and add a minus sign

36.

(included for
higher-level course)

FINDING THE DERIVATIVE
OF AN INVERSE FCT [function]
Let $\,f\,$ be a differentiable, 1-1 function.
Since $\,f\,$ is 1-1, it has an inverse $\,f^{-1}\,,$ and $$ f\bigl(f^{-1}(x)\bigr) = x\ \ \ \forall\ \ x\in\text{ran}(f) $$ Differentiating both sides wrt [with respect to] $\,x\,$
(and using the chain rule) gives: $$ f'\bigl(f^{-1}(x)\bigr)\cdot (f^{-1})'(x) = 1 $$ so that $$ \boxed{\ \underbrace{(f^{-1})'(x)}_{\text{Read: $\,f\,$ inverse, prime, of $\,x\,$}} = \frac 1{f'\bigl(f^{-1}(x)\bigr)}\ } $$ TWO METHODS:
FINDING THE DERIVATIVE
OF AN INVERSE FUNCTION
Let $\,f(x) = x^3\,,$ so $\,f^{-1}(x) = \root 3\of x = x^{1/3}\,.$
Find $(f^{-1})'(x)\,$ in 2 ways:

(1) Differentiating $\,f^{-1}\,$ directly: $$ \begin{align} (f^{-1})'(x) &= \frac d{dx} x^{1/3} = \frac 13x^{\frac 13 - 1}\cr &= \frac 13x^{-2/3} = \frac 1{3\root 3\of{x^2}} \end{align} $$ (2) Use the formula for $\,(f^{-1})'(x)\,$: $$ \begin{align} (f^{-1})'(x) &= \frac 1{f'\bigl(f^{-1}(x)\bigr)} \ \underbrace{=}_{f'(x) = 3x^2}\ \frac 1{3\bigl(f^{-1}(x)\bigr)^2}\cr\cr &\underbrace{=}_{f^{-1}(x) = x^{1/3}}\ \frac 1{3(x^{1/3})^2} = \frac 1{3\root 3\of{x^2}} \end{align} $$

37.

(included for
higher-level course)

WHAT DOES
$\,\displaystyle (f^{-1})'(x) = \frac 1{f'\bigl(f^{-1}(x)\bigr)}\,$
REALLY MEAN?

[left sketch: tangent line with slope $\,m = f'(a)\,$
to graph of $\,y = f(x)\,$ at point $\,(a,b)\,$
where $\,b = f(a)\,$ and $\,a = f^{-1}(b)\,$;
tangent line with slope $\,(f^{-1})'(b) = \frac 1m\,$ to graph of $\,y = f^{-1}(x)\,$ at point $\,(b,a)\,$]

[right sketch: $\,a\,$ gets sent by $\,f\,$ to $\,b\,$;
$\,b\,$ gets sent by $\,f^{-1}\,$ back to $\,a\,$]

Recall: $\,f\,$ and $\,f^{-1}\,$ are reflections about the line $\,y = x\,$ $$ \begin{align} (f^{-1})'(b) &= \frac 1{f'\bigl(f^{-1}(b)\bigr)}\cr &= \frac 1{f'(a)} = \frac 1m \end{align} $$
SUMMARY
For a differentiable, 1-1 fct [function] $\,f\,,$ with $\,m\ne 0\,,$ $$ \boxed{\ \substack{ \text{$\,(a,b)\,$ is a point}\cr \text{on the graph of $\,f\,$}\cr \text{with tangent line}\cr \text{of slope $\,m\,$} } \ } \iff \boxed{\ \substack{ \text{$\,(b,a)\,$ is a point}\cr \text{on the graph of $\,f^{-1}\,$}\cr \text{with tangent line}\cr \text{of slope $\,\frac 1m\,$} } \ } $$ In particular, the slopes of the tangent lines
are reciprocals!

38.

(included for
higher-level course)

DERIVATIVES OF THE INVERSE
TRIG [trigonometric] FCTS [functions]
$$ \begin{gather} \frac d{dx}\,\underbrace{\sin^{-1}x}_{\arcsin x} = \frac 1{\sqrt{1 - x^2}}\cr\cr \frac d{dx}\,\underbrace{\cos^{-1}x}_{\arccos x} = -\frac 1{\sqrt{1 - x^2}}\cr\cr \frac d{dx}\,\underbrace{\tan^{-1}x}_{\arctan x} = \frac 1{1 + x^2} \end{gather} $$ A TYPICAL DERIVATION
Recall: $$ \underbrace{y = \cos^{-1}x}_{ \substack{ \text{$\,\cos^{-1}x\,$ is the angle}\cr \text{between $\,0\,$ and $\,\pi\,$}\cr \text{whose cosine is $\,x\,$} } }\ \ \iff\ \ (\cos y = x\ \text{and}\ 0 \le y\le \pi) $$ [sketch showing the piece of the cosine curve on $\,[0,\pi]\,$ with point $\,(\cos^{-1}x,x)\,$]

$$ \begin{gather} \cos y = x\cr\cr (-\sin y)\frac{dy}{dx} = 1 \quad \substack{ \text{differentiate both sides}\cr \text{wrt [with respect to] $\,x\,$;}\cr \text{$\,y\,$ is a fct [function] of $\,x\,$} }\cr\cr \frac{dy}{dx} = \frac{-1}{\sin y} = \frac{-1}{\sqrt{1-x^2}} \end{gather} $$ [sketch explaining last equal sign:
triangle with angle $\,y = \cos^{-1}x\,$;
adjacent side $\,x\,,$ hypotenuse $\,1\,$

compute opposite side:
$\,x^2 + h^2 = 1\,,$ $\,h^2 = 1 - x^2\,,$ $\,h = \sqrt{1-x^2}\,$

$\,\cos y = \frac{\text{ADJ [adjacent]}}{\text{HYP [hypotenuse]}} = \frac{x}1 = x$]

39.

Implicit Differentiation

DIFFERENTIATING VARIABLE STUFF
TO VARIABLE POWERS;
THE LOG TRICK
For all $\,x > 0\,,$ $\,x = {\text{e}}^{\ln x}\,.$
[sketch: $\,x\,$ into $\,\ln\,$ box with output $\,\ln x\,$;
$\,\ln x\,$ into $\,{\text{e}}^x\,$ box with output $\,{\text{e}}^{\ln x} = x\,$;
$\,\ln\,$ and $\,{\text{e}}^x\,$ are inverses; undo each other]

Also, $\,\ln A^B = B\ln A\,.$
Thus, $$ \begin{align} f(x)^{g(x)} &= {\text{e}}^{\ln (f(x)^{g(x)})}\cr &= {\text{e}}^{g(x)\ln f(x)} \end{align} $$ and we know that $\,\frac{d}{dx} {\text{e}}^{f(x)} = {\text{e}}^{f(x)}\cdot f'(x)\,$!

EXAMPLE
Differentiate $\,x^x\,.$
SOLN [solution]: $$x^x = {\text{e}}^{\ln(x^x)} = {\text{e}}^{x\ln x}\,;$$ $$ \begin{align} \frac d{dx} x^x &= \frac d{dx} {\text{e}}^{x\ln x}\cr &= {\text{e}}^{x\ln x}\cdot \frac d{dx}(x\ln x)\cr &= x^x(x\cdot \frac 1x + \ln x)\cr &= x^x(1 + \ln x) \end{align} $$

40.

(included for
higher-level course)

L'HOPITAL'S RULE
... helps us investigate indeterminate forms!

Spse [suppose] $\,f\,$ and $\,g\,$ are differentiable, and $\,g'(x)\ne 0\,$ near $\,a\,$ (except possibly at $\,a\,$). Spse that $\,\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\,$ is a $\,\frac 00\,$ or $\,\frac{\pm\infty}{\pm\infty}\,$ indeterminate form. Then, $$ \begin{gather}\text{IF } \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\ \text{ exists, or equals $\,\pm\infty\,,$ }\cr \text{THEN } \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)} \end{gather} $$ MOTIVATION, “$\,\frac 00\,$” CASE
[sketch: graphs of both $\,f\,$ and $\,g\,$ cross the $x$-axis at $\,a\,$]
$\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\,$ is “$\,\frac 00\,$” indeterminate form

Zoom in at ‘$\,a\,$’:   for $\,x\,$ near $\,a\,,$ $$ \begin{align} \frac{f(x)}{g(x)} &\cong \frac{f'(a)(x-a)}{g'(a)(x-a)}\cr &= \frac{f'(a)}{g'(a)} \end{align} $$ [sketch: the tangent lines to $\,f\,$ and $\,g\,$ at the point $\,(a,0)\,$ have equations $\,y = f'(a)(x-a)\,$ and $\,y = g'(a)(x-a)\,$]

41.

(included for
higher-level course)

BASIC USE OF L'HOPITAL'S RULE
EX [example]:   $\displaystyle\lim_{x\rightarrow 3}\frac{x-3}{x^2-9} \ \underset{\text{l'H}} {\overset{\text{“$\frac 00$”}}{=}}\ \lim_{x\rightarrow 3}\frac 1{2x} = \frac 16$

EX:   $\displaystyle \lim_{x\rightarrow\infty}\frac{{\text{e}}^x}{x^2} \ \underset{\text{l'H}} {\overset{\text{“$\frac \infty\infty$”}}{=}}\ \lim_{x\rightarrow\infty}\frac{{\text{e}}^x}{2x} \ \underset{\text{l'H}} {\overset{\text{“$\frac \infty\infty$”}}{=}}\ \lim_{x\rightarrow\infty}\frac{{\text{e}}^x}2 = \infty $
$$ \begin{align} \text{EX:}\ \ \lim_{x\rightarrow 0^+} x\ln x \overset{\text{“$\,0(-\infty)\,$”}}{=} \lim_{x\rightarrow 0^+}\frac{\ln x}{1/x} &\ \underset{\text{l'H}} {\overset{\text{“$\frac{-\infty}\infty$”}}{=}}\ \lim_{x\rightarrow 0^+}\frac{\frac 1x}{\frac{-1}{x^2}}\cr &= \lim_{x\rightarrow 0^+} (-x) = 0 \end{align} $$ [technique for $\,(0)(\pm\infty)\,$: rename $\,x\ln x\,$ as $\,\frac{\ln x}{1/x}\,$ to get a form that l'Hopital's rule handles]

MORE ADVANCED USE
OF L'HOPITAL'S RULE
$$ \begin{align} \lim_{x\rightarrow\infty} 7x{\text{e}}^{1/x} - 7x &\overset{\text{“$\,\infty-\infty\,$”}}{=} \lim_{x\rightarrow\infty} 7x({\text{e}}^{1/x} - 1)\cr\cr &\overset{\text{“$\,(\infty)(0)\,$”}}{=} \lim_{x\rightarrow\infty} \frac{{\text{e}}^{1/x}-1}{(1/(7x))} \ \underset{\text{l'H}} {\overset{\text{“$\frac 00$”}}{=}}\ \lim_{(x\rightarrow\infty)}\frac{{\text{e}}^{1/x}(-x^{-2})}{-(7x)^{-2}\cdot 7}\cr\cr &= \lim_{x\rightarrow\infty}\frac{{\text{e}}^{1/x}(7x)^2}{x^2\cdot 7} = \lim_{x\rightarrow\infty}\frac{49{\text{e}}^{1/x}}7\cr\cr &= 7 \end{align} $$

42.

the Test Point Method
Logical Equivalences

WHERE CAN A FUNCTION
CHANGE ITS SIGN?
... from positive to negative,
or from negative to positive?
[sketches: $\,x\,$]
ANS [answer]: where $\,f(x) = 0\,,$
or at a break in the graph

NEGATING ‘AND’ & ‘OR’ SENTENCES
$$ \begin{align} \text{not}(A\text{ or } B) \ \ &\iff\ \ (\text{not }A) \text{ and } (\text{not }B)\cr \text{not}(A\text{ and } B) \ \ &\iff\ \ (\text{not }A) \text{ or } (\text{not }B) \end{align} $$ [left sketch: $\text{not}(A \text{ or }B)$ ]
[right sketch: $(\text{not }A)\text{ and }(\text{not }B)$ ]

$A$	$B$	$A\text{ or }B$	$\text{not}(A\text{ or }B)$	$\text{not }A$	$\text{not }B$	$(\text{not }A) \text{ and } (\text{not }B)$
T	T	T	F	F	F	F
T	F	T	F	F	T	F
F	T	T	F	T	F	F
F	F	F	T	T	T	T

[Compare the two indicated columns—they're the same!]

43ab.

Graphing Functions

INFO [information] GIVEN BY $\,f\,,$ $\,f'\,,$ $\,f''\,$
$$ \begin{gather} f > 0\ \ \iff\ \ \text{graph of $\,f\,$ above $x$-axis}\cr\cr f \lt 0\ \ \iff\ \ \text{graph of $\,f\,$ below $x$-axis}\cr\cr \overbrace{f' > 0}^{\text{slopes of $\,f\,$ positive}}\ \ \iff\ \ \text{graph of $\,f\,$ increasing [going uphill from left to right]}\cr\cr \overbrace{f' \lt 0}^{\text{slopes of $\,f\,$ negative}}\ \ \iff\ \ \text{graph of $\,f\,$ decreasing [going downhill from left to right]}\cr\cr\cr f'' > 0\ \ \iff\ \ \underbrace{\text{$\,f'\,$ increasing}}_{\text{slopes of $\,f\,$ increasing}} \ \ \iff\ \ \text{$\,f\,$ concave up [holding water]}\cr\cr f'' \lt 0\ \ \iff\ \ \underbrace{\text{$\,f'\,$ decreasing}}_{\text{slopes of $\,f\,$ decreasing}} \ \ \iff\ \ \text{$\,f\,$ concave down [shedding water]}\cr \end{gather} $$ A BASIC SIGN ANALYSIS OF A FUNCTION
Let $\,f(x) = x{\text{e}}^x\,$;   find the sign of $\,f\,,$ $\,f'\,,$ $\,f''\,$
$$ \begin{gather} f(x) = 0\ \ \iff\ \ x = 0\ \ ({\text{e}}^x > 0\ \ \forall \ x)\cr\cr f'(x) = x{\text{e}}^x + {\text{e}}^x = {\text{e}}^x(x + 1)\,;\ \ f'(x) = 0\ \ \iff\ \ x = -1\cr\cr f''(x) = {\text{e}}^x\cdot 1 + (x+1){\text{e}}^x = {\text{e}}^x(1 + x + 1) = {\text{e}}^x(x+2)\,;\cr f''(x) = 0\ \ \iff\ \ x = -2 \end{gather} $$ [mark ‘z’ as the mark for zero]
[z at $\,-2\,$: negative to left, positive to right] Sign of $\,f''\,$
[z at $\,-1\,$: negative to left, positive to right] Sign of $\,f'\,$
[z at $\,0\,$: negative to left, positive to right] Sign of $\,f\,$

Graph of $\,f\,$:
[$\,(-\infty,-2)\,$: neg [negative], dec [decreasing], concave down]
[$\,(-2,-1)\,$: neg, dec, concave up]
[$\,(-1,0)\,$: neg, inc [increasing], concave up]
[$\,(0,\infty)\,$: pos [positive], inc, concave up]

43cd.

If/Then Sentences
Contrapositive and Converse

IMPLICATIONS
(AND EQUIVALENT SENTENCES)
If $\,A\,,$ then $\,B\,$
$A\Rightarrow B$     ($\,A\,$ implies $\,B\,$)
$B\,,$ whenever $\,A\,$
Whenever $\,A\,,$ $\,B\,$
$A\,$ is sufficient for $\,B\,$

$A\,$ is the hypothesis of the implication;
$B\,$ is the conclusion of the implication

$A$	$B$	If $\,A\,,$ then $\,B\,$
T	T	T
T	F	F
F	T	T
F	F	T

[first row]: To prove that an implication is true, we must show than whenever the hypothesis is T [true], so is the conclusion
[last two rows]: “vacuously true”

THE CONTRAPOSITIVE OF AN IMPLICATION
The contrapositive of  “$A\Rightarrow B\,$” is $\,(\text{not } B) \Rightarrow (\text{not } A)\,$

$A$	$B$	$A\Rightarrow B$	$\text{not }B$	$\text{not }A$	$(\text{not }B) \Rightarrow (\text{not }A)$
T	T	T	F	F	T
T	F	F	T	F	F
F	T	T	F	T	T
F	F	T	T	T	T

[compare third and last columns]: Same!!
An implication is equivalent to its contrapositive!

44.

Critical Points

WHERE CAN A FCT [function]
HAVE A LOCAL MAX/MIN?
At a place where there is a horizontal tangent line
$\,f'(c) = 0\,$

At a place where the fct is not differentiable
$\,f'(c)\,$ DNE (Does Not Exist)

At an endpoint of the domain of $\,f\,$
endpt [endpoint] of $\,\text{dom } f\,$

[other side of index card]
If $\,f\,$ has a local max/min at $\,c\,,$
then $\bigl( f'(c) = 0\ \ \ \text{or}\ \ \ f'(c) \text{ DNE}\ \ \ \text{or}\ \ \ c\, \text{ is an endpt of $\text{dom}(f)\,$}\bigr)$

Recall: $$\text{If } P, \text{ then } Q\ \ \iff\ \ \underbrace{\text{If } (\text{not }Q), \text{ then } (\text{not }P)}_{\text{contrapositive}} $$ If $\bigl( f'(c)\ne 0\ \ \ \text{and}\ \ \ f \text{ IS differentiable at $\,c\,$}\ \ \ \text{and}\ \ \ c \,\text{ is NOT an endpt of $\,\text{dom}(f)\,$} \bigr) $
then
$\,f\,$ does not have a local max/min at $\,c\,$

45.

Critical Points

CRITICAL POINTS;
CRITICAL NUMBERS
DEFN [definition]: $\,c\,$ is a critical number for $\,f\,$
iff [if and only if]
$\,c\,$ is a # [number] in the domain of $\,f\,$ for which $\,f'(c) = 0\,,$ or $\,f'(c)\,$ DNE [Does Not Exist],
or $\,c\,$ is an endpt [endpoint] of $\,\text{dom}(f)\,.$

DEFN: $\,(c,f(c))\,$ is a critical point iff $\,c\,$ is a critical #.

The critical points are the candidates
for where local max/min occur!

CAREFUL!
A critical point does not have to be a max or min!!

[sketch: horizontal tangent line at $\,c\,,$
increasing to both left and right]
$\,f'(c) = 0\,,$ but no max/min

[sketch: kink at $\,c\,,$
increasing to both left and right]
$\,f'(c)\,$ DNE, but no max/min

We're not guaranteed a max/min at critical
pts [points]—they're just the candidates.

46.

Inflection Points

WHERE CAN A FUNCTION HAVE
AN INFLECTION POINT?
An inflection point is a place where the concavity changes—from concave up to concave down, or concave down to concave up.

(An inflection pt [point] can't occur at
an endpt [endpoint]—you need to know
what happens on both sides.)

[left sketch: inflection point at $\,c\,$:
concave down to left, concave up to right,
$f''(c) = 0\,$]

[right sketch: inflection point at $\,c\,$:
concave up to left, concave down to right,
$f''(c)\,$ DNE [Does Not Exist]]

[other side of card]
If $\,f\,$ has an inflection pt at $\,c\,,$
then ($\,f''(c) = 0\,$ or $\,f''(c)\,$ DNE ).

If ($\,f''(c)\ne 0\,$ and $\,f''\,$ does exist [at $\,c\,$] )
then $\,f\,$ does not have an inflection pt at $\,c\,.$

So, places where $\,f''(c) = 0\,,$ $\,f''(c)\,$ DNE
are the candidates for inflection pts.

[left sketch]: $\,y = x^4\,,$
$\,y'' = 0\,$ at $\,x = 0\,$;
not an inflection pt

[right sketch]: $f''(c)\,$ DNE;
not an inflection pt

47.

Graphing Functions

AN ALGORITHM FOR
THOROUGH FUNCTION ANALYSIS

• find $\,\text{dom}(f)$
• find first derivative, $\,f'$
• find critical pts [points]: $f' = 0\,,$
  $\,\overbrace{f' \text{ DNE [Does Not Exist]}}^{\text{but $\,f\,$ does!}}\,,$
  endpts [endpoints] of $\,\text{dom}(f)\,$
• find second derivative, $\,f''$
• find candidates for inflection pts:
  $\,f'' = 0\,,$ $\,f''\,$ DNE
• create a derivative chart (see 47b)
• characterize local max/min, inflection pts;
  find $y$-values
• determine end behavior, as needed
• sketch graph
• identify global max/min

EXAMPLE:
THOROUGH FUNCTION ANALYSIS
$f(x) = x{\text{e}}^x\,$; see card 43 for derivatives

$x$	$(-\infty,-2)$	$-2$	$(-2,-1)$	$-1$	$(-1,\infty)$
$f(x)$	[arrow showing decreasing, concave down]	INFL [inflection pt] $f(-2) = \frac{-2}{{\text{e}}^2}$	[arrow showing decreasing, concave up]	local min $f(-1) = -\frac 1{\text{e}}$	[arrow showing increasing, concave up]
$f'(x)$	$-$ $\searrow$	$-$	$-$ $\nearrow$	$0$	$+$ $\nearrow$
$f''(x)$	$-$	$0$	$+$	$+$	$+$

[notes on side]:
$x$-int [intercept]: $\,x{\text{e}}^x = 0\,$ iff $\,x = 0$
$y$-int: $\,f(0) = 0\cdot{\text{e}}^0 = 0$
As $\,x\rightarrow\infty\,,$ $\,y\rightarrow\infty$

$\displaystyle\lim_{x\rightarrow-\infty} x{\text{e}}^x = \lim_{x\rightarrow -\infty} \frac{x}{{\text{e}}^{-x}} \underset{\strut\text{l'H}}{\overset{\text{“$\frac{-\infty}{\infty}$”}}{=}} \lim_{x\rightarrow -\infty} \frac 1{-{\text{e}}^{-x}} = 0 $
[l'H = l'Hopital's rule]

48.

Local Max/Min
the Second Derivative

ARE CRITICAL PTS [points]
ACTUALLY MAX OR MIN?
TWO TESTS: 1^st and 2^nd DERIVATIVE TESTS

THE SECOND DERIVATIVE TEST:
Spse [suppose] $\,f''\,$ is continuous near $\,c\,.$
If $\,f'(c) = 0\,$ and $\,f''(c) > 0\,,$
then $\,f\,$ has a local min @ [at] $\,c\,.$
[top sketch: concave up, horizontal tangent line at $\,c\,$]

If $\,f'(c) = 0\,$ and $\,f''(c) \lt 0\,,$
then $\,f\,$ has a local max @ $\,c\,.$
[bottom sketch: concave down, horizontal tangent line at $\,c\,$]

The $\underbrace{\text{test fails}}_{\text{is inconclusive}}$ if $\,f''(c) = 0\,$:
might be a max, min, or neither.

THE FIRST DERIVATIVE TEST
Spse $\,c\,$ is a critical # [number] for $\,f\,,$
and $\,f\,$ is continuous at $\,c\,.$

[all sketches have a number line showing $\,c\,,$
the SIGN OF $\,f'\,,$ the GRAPH OF $\,f\,,$
and the conclusion (local max, local min, or neither)]

The 1^st and 2^nd deriv [derivative] tests
are used to decide if there are max/min
at some (but not all) critical points.

[side of card]: Why continuity needed?
[sketch: derivative is positive to left of $\,c\,$
and negative to right of $\,c\,,$
but there is NOT a local max at $\,c\,$]

49.

Continuity

OPEN/CLOSED INTERVALS
An interval is open when
it doesn't include any endpoint:
e.g., $\,(a,b)\,,$ $\,(a,\infty)\,$
[sketch showing hollow dots at $\,a\,$ and $\,b\,,$
everything shaded between]

Every pt [point] in an open interval has some “space” to both the right and left, that is inside the interval.

An interval is closed when its complement is open;
closed intervals include their endpt(s):
e.g., $\,(-\infty,b]\,,$ $\,[a,b]\,$

Note: $\,(a,b]\,$ is not open, and not closed!
These words are used differently in English and math!

[side of card]: Recall: If the universal set is $\,\Bbb R\,,$
and $\,A\,$ is a subset of $\,\Bbb R\,,$ then $$ \begin{gather} \sim\kern-3pt A = \Bbb R\backslash A = \Bbb R - A = A^c = \overline{A}\cr = \text{ the complement of $\,A\,$} = \{x\in\Bbb R\ |\ x\notin A\} \end{gather} $$ BOUNDED SUBSETS OF $\,\Bbb R$
DEFN [definition]: A set $\,S\,$ of real #s [numbers] is bounded above if there is a real # $\,k\,$ s.t. [such that] $\,k\ge s\ \ \forall\ s\in S\,.$
($\,k\,$ is called an upper bound [of $\,S\,$])

Also, $\,S\,$ is bounded below if there is a real # $\,k\,$ s.t. $\,k\le s\ \ \forall\ s\in S\,.$
($\,k\,$ is a lower bound [of $\,S\,$])

$\,S\,$ is bounded if it has both upper and lower bounds.

Note: $\,[a,b]\,$ is a bounded, closed interval.
[where $\,a \lt b\,$]

50.

The Max-Min Theorem

REVIEW OF ABSOLUTE/GLOBAL MAX/MIN
When an absolute max/min exists, then it is unique. [sketch showing ‘all $\,x\in\text{dom}(f)\,$’
going into the $\,f\,$ box]
Is there a greatest # [number] in this pile?
If so, it's the absolute max.
Is there a least # in this pile?
If so, it's the absolute min.

However, an abs [absolute] max/min can occur
at many different input(s) ($x$-values).

Note: A fct [function] may not have an abs max/min.
[sketch showing two functions with
no absolute max and no absolute min]

THE EXTREME VALUE THEOREM
[also called The MAX/MIN THEOREM]

THM [theorem]: A continuous function on a bounded closed interval has a global max and a global min.

[sketch: $\,(a,m)\,$ is a global min point;
$\,(c,M)\,$ and $\,(b,M)\,$ are both global max points;
graph is labeled $\,f\,$]

Global max is $\,M\,,$ at $\,x = c\,$ and $\,x = b\,$
Global min is $\,\,m\,,$ at $\,x = a\,$

Note: Take away either hypothesis,
and the conclusion may fail:
[sketch: a function (not continuous) on $\,[a,b]\,$
with no global min]
[sketch: a (continuous) function on $\,(a,b]\,$
with no global max]

51.

Local Max/Min

OPTIMIZATION PROBLEMS
(FINDING MAX/MIN)

• Determine the fct [function] to be maximized/minimized. It must be a fct of only one variable. Call it $\,f\,.$
• Find $\,\text{dom}(f)\,$; is $\,f\,$ continuous? If $\,f\,$ is continuous on $\,[a,b]\,,$ then it must have both a global max and a global min.
• Find all critical pts [points]—any max/min must occur at one of these pts. Organize your results in a table (51b).
• Decide if each candidate is a global/local max/min, or not; use the 1^st and 2^nd derivative tests, as needed.

EXAMPLE
Find the absolute max/min of $\displaystyle\,f(x) = \frac{\ln x}x\,$ on $\,[1,3]\,.$
Note: $\,f\,$ is cont [continuous] on $\,[1,3]\,$; it is guaranteed to have both a global max and a global min. $$ f'(x) = \frac{x\cdot\frac 1x - \ln x}{x^2} = \frac{1 - \ln x}{x^2} \overset{\text{set}}{=} 0\ ; $$ the only way a fraction can $\,= 0\,$
is for its numerator to $\,= 0\,$; $$ 1 - \ln x = 0\ \ \iff\ \ 1 = \ln x\ \ \iff\ \ x = \text{e} $$ Use a table like this to summarize work:

$c$	$f(c)$	Why a critical pt [point]?	Conclusion
$1$	$\frac{\ln 1}1 = 0$	endpt [endpoint] of $\text{dom}(f)$	abs min $= 0$
$3$	$\frac{\ln 3}3 \approx 0.366$	endpt of $\text{dom}(f)$
$\text{e}$	$\frac{\ln \text{e}}{\text{e}} \approx 0.368$	$f' = 0$	abs max $= \frac 1{\text{e}}$

[on left side]: The greatest value must be the abs max; least value must be the abs min!!

52.

Local Max/Min

A FIRST OPTIMIZATION PROBLEM
A rancher needs to fence a rectangular area with 2400 ft [feet] of fence; one side is along a river, so won't need fence. Find the maximum area that can be enclosed.
[sketch: height $\,x\,,$ width $\,y\,$] $$ \begin{gather} 2x + y = 2400\cr y = 2400 - 2x\cr A = xy = x(2400 - 2x)\,; \ \ \text{dom}(A) = [0,1200]\cr A'(x) = 2400 - 4x \overset{\text{set}}{=} 0\cr 2400 = 4x\cr x = 600\cr A''(x) = -4\,; \ \ \underbrace{A''(600) \lt 0}_{\text{2nd Deriv [derivative] Test}} \end{gather} $$ [sketch: graph of $\,A\,$: concave down parabola,
vertex $\,(600\,,\,720{,}000)\,,$ $x$-intercept at $\,1200\,$]

[on side]: ANS: The max area enclosed
is 720,000 sq ft [square feet]

YOU TRY IT!!
(1) Same problem—no looking!

(2) Same problem, but no river on one side.
[upside-down: ANS: 360,000 sq ft]

(3) A rancher wants to fence an area of 3,000,000 sq ft in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence the rancher can use?
[upside down: ANS: $\,\approx \overbrace{8485.28}^{6000\sqrt 2}\,$ ft]

53ab.

Local Max/Min

AN OPTIMIZATION PROBLEM
Find the maximum area of a rectangle inscribed in a semi-circle of radius $\,R\,.$
[sketch: $\,x\,,$ $\,R\,$]
$$ \begin{gather} x^2 + y^2 = R^2\cr y = \sqrt{R^2 - x^2} \quad \text{(upper half)}\cr A(x) = 2x\sqrt{R^2 - x^2}\cr \text{dom}(A) = [0,R]\cr \end{gather} $$ $$ \begin{align} A'(x) &= 2x\cdot \frac 12(R^2 - x^2)^{-1/2}\cdot(-2x) + \sqrt{R^2 - x^2}\cdot 2\cr &\underset{\text{[algebra]}}{\overset{\text{alg}}{=}} \frac{2(R^2 - 2x^2)}{\sqrt{R^2 - x^2}} \overset{\text{set}} = 0 \end{align} $$ ... CONTINUED
$$ \begin{gather} R^2 - 2x^2 = 0\cr 2x^2 = R^2\cr x^2 = \frac{R^2}2\cr x = \frac{R}{\sqrt 2} \quad \text{(take the positive soln [solution])} \end{gather} $$ $$ \begin{align} A\bigl(\frac R{\sqrt 2}\bigr) &= 2\bigl(\frac{R}{\sqrt 2}\bigr)\sqrt{R^2 - \frac{R^2}2}\cr &\overset{\text{alg}}{=} R^2\,; \end{align} $$ this is the max since $\,0 = A(0) = A(R)\,$

The max area is $\,R^2\,$

You try it: A rectangle is inscribed with its base on the $x$-axis and its upper corners on the parabola $\,y = 11 - x^2\,.$ What are the dimensions of such a rectangle with the greatest area?
[upside-down: ANS: width $\,= 2\sqrt{\frac{11}3} \approx 3.83\,$;
height $\,= \frac{22}3 \approx 7.33\,$]

53cd.

Local Max/Min

A USEFUL OBSERVATION
[sequence of graphs: $\,f\,$ on top, $\,f^2\,$ on bottom;
outputs change, but max/min are preserved for $\,f > 0\,$]

If $\,\boxed{\,f > 0\,}\,$ and $\,f\,$ has a max or min at $\,c\,,$
then $\,f^2\,$ also has a max/min at $\,c\,.$

USING THE OBSERVATION
So, if we ever want to maximize/minimize
something of the form $\,\sqrt{f(x)}\,$ (which is $\,\ge 0\,$),
then we can instead optimize $\,f(x)\,,$ which is easier!!

54.

Local Max/Min

AN OPTIMIZATION PROBLEM
Find the point on the parabola $\,y^2 = 2x\,$
that is closest to the point $\,(1,4)\,.$ $$ \begin{align} \text{distance from $\,(x,y)\,$ to $\,(1,4)\,$} &= \sqrt{(x-1)^2 + (y-4)^2}\cr &= \sqrt{(\frac{y^2}2 - 1)^2 + (y-4)^2}\ \ ; \end{align} $$

Let

$$ f(y) = \frac{y^2}2 - 1)^2 + (y-4)^2\ ; $$

so

$$ f'(y) = \underbrace{2(\frac{y^2}2-1)\cdot y + 2(y-4)}_{= y^3 - 8} \ \overset{\text{set}}{=}\ 0 $$ [second side of card]
$$ \begin{gather} (y^2-2)y + 2(y-4) = 0\cr y^3 - 2y + 2y - 8 = 0\cr y^3 = 8\cr y = 2\ ; \ \text{ so }\ x = \frac{y^2}2 = 2 \end{gather} $$ [number line: $\,2\,,$ negative to left, positive to right,
labeled ‘Sign of $\,f'(y)\,$’]

By 1^st Derivative Test, $\,(2,2)\,$ is a local min
(which is also a global min).
Thus, the pt [point] $\,(2,2)\,$ is closest to $\,(1,4)\,.$

You try it:
Find the pt on the line $\,-2x + 3y + 3 = 0\,$
that is closest to the pt $\,(-3,1)\,.$
[upside down: ANS: $\,x = -\frac{15}{13}\,,$ $\,y = -\frac{23}{13}\,$]

55.

Implicit Differentiation

$\underbrace{\text{EXPLICIT}}_{\text{“showing”}}\,$ versus $\underbrace{\text{IMPLICIT}}_{\text{“hidden”}}\,$
When $\,y\,$ is given explicitly as a fct [function] of $\,x\,$
($\,y = f(x)\,$), then the fct relationship
between $\,x\,$ and $\,y\,$ is clear.

When $\,y\,$ is given implicitly in terms of $\,x\,,$
then $\,y\,$ may only be a fct of $\,x\,$ locally.

EX [example]: $\,x^2 + y^2 = 9\,$
[graph of circle, showing a value of $\,x\,$ with TWO values of $\,y\,$ and corresponding tangent lines]
We can still talk about slopes of tangent lines at most pts, but we usually need to know both $\,x\,$ and $\,y\,$ values!

IMPLICIT DIFFERENTIATION
... is a technique to find $\,\frac{dy}{dx}\,,$
when $\,y\,$ is given implicitly in terms of $\,x\,.$

EX: Find slope of tangent line to $\,x^2 + y^2 = 9\,$
at $\,(2,\sqrt 5)\,.$

Differentiate both sides wrt [with respect to] $\,x\,$; remember that $\,y\,$ is (at least locally!) a fct of $\,x\,$: $$ \begin{gather} x^2 + y^2 = 9\cr 2x + 2y\frac{dy}{dx} = 0 \end{gather} $$ Formula depends on both $\,x\,$ and $\,y\,$: $$ \begin{gather} \frac{dy}{dx} = \frac{-2x}{2y} = -\frac xy\cr \frac{dy}{dx}\bigg|_{\substack{x = 2\cr y=\sqrt 5}} = \frac{-2}{\sqrt 5} \end{gather} $$

56.

(included for
higher-level course)

RELATED RATE PROBLEMS
IDEA: Compute the rate of change of one quantity
in terms of the rate of change of another quantity
(which may be more easily measured).

PROCEDURE: Find an equation
that relates the two quantities
and use the Chain Rule
to differentiate both sides with respect to time.

FALLING LADDER PROBLEM
A 10 ft ladder rests against a vertical wall.
If the bottom slides away from the wall at 1 ft/sec,
how fast is the top sliding down the wall,
when the bottom is 6 ft from the wall?

[sketch: 10 ft ladder, top is distance $\,y\,$ from floor,
bottom is distance $\,x\,$ from wall]

SOLN [solution]: $$ \begin{gather} x^2 + y^2 = 100 \quad \text{(relate quantities)}\cr 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\quad \text{(differentiate wrt [with respect to] $\,t\,$)}\cr \frac{dy}{dt} = \frac{-2x\frac{dx}{dt}}{2y} = -\frac xy\frac{dx}{dt} \ \ \text{ and }\ \ \frac{dx}{dt} = 1 \end{gather} $$ When bottom is 6 ft from wall: $$ 6^2 + y^2 = 10^2\,;\ \ y = 8 $$

so

$$ \frac{dy}{dt}\biggl|_{\substack{x = 6\cr y = 8}} = \frac{-6}8\cdot 1 = \frac{-3}4 \text{ ft/sec} $$

57.

(included for
higher-level course)

IS THE CAR SPEEDING?
A police officer with a radar gun is by the side of the road. The gun measures a car approaching at 50 mph; at that moment the angle between the road and the line of sight between the office and the car is 30°.
Is the car violating the 55 mph speed limit?
[sketch: Officer is (perpendicular) distance $\,a\,$ from road
and (diagonal) distance $\,y\,$ from car;
bottom leg of triangle is $\,x\,,$ angle $\,\theta\,$]

Note: When $\,\theta = 30^\circ\,,$ $$\frac{dy}{dt} = -50\,;$$ want to know $\,\frac{dx}{dt}\,.$

[other side of index card]
$$ \begin{gather} a^2 + x^2 = y^2\cr\cr 2x\frac{dx}{dt} = 2y\frac{dy}{dt}\cr\cr \frac{dx}{dt} = \frac yx\frac{dy}{dt} \end{gather} $$ When $\,\theta = 30^\circ\,,$ $\,\cos 30^\circ = \frac xy\,,$ so $$ \frac yx = \frac 1{\cos 30^\circ} = \frac 1{\sqrt 3/2} = \frac 2{\sqrt 3} $$

so

$$ \frac{dx}{dt}\bigg|_{\theta = 30^\circ} = \frac 2{\sqrt 3}(-50) = \frac{-100}{\sqrt 3} \simeq \underset{\substack{\uparrow\cr x\cr\text{is}\cr\text{decreasing}}}{-}\kern{-10pt}57.74 \text{ mph} $$ The car is speeding!

58.

the Mean Value Theorem

THE MEAN VALUE THEOREM
THM [theorem]: Suppose $\,f\,$ is differentiable on $\,(a,b)\,$
and continuous on $\,[a,b]\,.$
Then there exists $\,c\in (a,b)\,$ for which $$ f'(c) = \frac{f(b)-f(a)}{b - a} $$ [sketch: graph of function $\,f\,$ on interval $\,[a,b]\,$;
tangent line at $\,c\,$ has slope $\,m = f'(c)\,$;
slope of secant line between endpoints of interval is $m = \frac{f(b)-f(a)}{b-a}$]

THE MEAN VALUE THEOREM (MVT)
Guarantees a place where the instantaneous rate of change is the same as the average rate of change on a given interval.

INTUITION: Whenever you're given info [information] about a derivative, and want info about the function itself—think MVT!!

59.

the Mean Value Theorem

A MEAN VALUE THEOREM EXAMPLE
Suppose $\,1\le f'(x)\le 4\ \ \forall\ \ x\in [2,5]\,.$
Show that $\,3\le f(5)-f(2)\le 12\,.$

SOLN [solution]: By the MVT [Mean Value Theorem], $\,\exists\ \ c\in (2,5)\,$ for which $$ f'(c) = \frac{f(5)-f(2)}{5-2}\ , $$ so $\, 3f'(c) = f(5) - f(2)\,.$
But $\,1\le f'(c)\le 4\,$ (Why?)
So $\,3\le 3f'(c)\le 12\,.$
Thus, $\,3\le f(5) - f(2)\le 12\,.$

A MEAN VALUE THEOREM EXAMPLE
Consider the function $\,f(x) = 5-3x^2\,$ on $\,[-3,7]\,.$
(a) Find the average ROC [rate of change]
of $\,f\,$ on $\,[-3,7]\,$:
$$ \begin{align} \frac{f(7)-f(-3)}{7-(-3)} &= \frac{(5-3\cdot7^2) - (5-3(-3)^2)}{10}\cr &= -12 \end{align} $$ (b) By the MVT, $\,\exists\ \ c\in (-3,7)\,$ for which $\,f'(c)\,$ equals this AVROC [average rate of change]. Find $\,c\,$:
$$ \begin{gather} f'(x) = -6x\cr -6c = -12\cr c = 2 \end{gather} $$

60.

the Definite Integral

BOUNDED FUNCTIONS

DEFN [definition]: A function $\,f\,$ is bounded on a set $\,A\,$ if the output values of $\,f\,$ on $\,A\,$ are contained in a bounded interval.

[sketch of $\,y = \frac 1x\,$]
$\,y = \frac 1x\,$ is not bounded on $\,(0,1]\,$
since it takes on arbitrarily large values.

Note: Every continuous function on an interval $\,\underbrace{[a,b]}_{\text{closed and bounded}}\,$ is bounded on that interval.

[sketch of a continuous function
on a closed and bounded interval]

THE DEFINITE INTEGRAL OF A FUNCTION
DEFN: Assume $\,f\,$ is bounded on $\,[a,b]\,.$
The definite integral of $\,f\,$ from $\,a\,$ to $\,b\,$
is the signed area of the region
between the graph of the function and the $x$-axis.
Area above the $x$-axis is positive;
area below the $x$-axis is negative.
The definite integral is denoted by $$\int_a^b f(x)\,dx$$ or, more simply: $$\int_a^b f$$ [sketch, showing positive and negative areas]

61.

the Definite Integral

A DEFINITE INTEGRAL
MAY OR MAY NOT EXIST

• if $\,\int_a^b f\,$ exists, we say the function $\,f\,$ is integrable on $\,[a,b]\,$
• continuous functions on $\,[a,b]\,$ are integrable
• a function with a finite # [number] of removable or jump discontinuities is integrable
  [sketches of removable and jump discontinuities]

NOTATION FOR THE DEFINITE INTEGRAL
$$ \int_a^b \underbrace{f(x)}_{\text{integrand}}\,dx $$ [the integrand is the fct [function] being integrated;
$\,a\,$ and $\,b\,$ are the limits of integration;
$\,a\,$ is the lower limit ;
$\,b\,$ is the upper limit]

Think of $\,\int_a^b\,$ and $\,dx\,$ as a pair;
the fct being integrated gets smushed between!
The process of finding $\,\underbrace{\int_a^b f(x)\,dx}_{\text{a number}}\,$
is called integration.

62.

the Definite Integral

DUMMY VARIABLES
IN DEFINITE INTEGRALS
$$ \int_a^b f(x)\,dx = \int_a^b f(t)\,dt = \int_a^b f(w)\,dw $$ [first integral: dummy variable is $\,x\,$;
does not affect the value of the expression]

[second integral: dummy variable is $\,t\,$]

[third integral: dummy variable is $\,w\,$;
$\,a\,,$ $\,b\,,$ and $\,f\,$ are NOT dummy variables!
Changing the fct [function] and/or the limits
can change the value of the integral!]

SIMPLE DEFINITE INTEGRAL EXAMPLES
$$ \int_0^2 3\,dx = 6 $$ [sketch showing graph of $\,y = 3\,$ on $\,[0,2]\,$;
$\,\text{AREA} = 6\,$]

$$ \int_0^2 (-3)\,dx = -6 $$ [sketch showing graph of $\,y = -3\,$ on $\,[0,2]\,$;
$\,\text{SIGNED AREA} = -6\,$]

$$ \int_{-2}^2 x\,dx = 0 $$ [sketch showing graph of $\,y = x\,$ on $\,[-2,2]\,$;
same amt [amount] of area above and below;
cancels out!]

63.

the Definite Integral

PROPERTIES OF
THE DEFINITE INTEGRAL
DEFN [definition]: If $\,a \lt b\,,$ we define $$ \int_b^a f := -\int_a^b f $$ “Integrating backwards introduces a minus sign”

If $\,f\,$ is integrable,
then for all real #s [numbers] $\,a\,,$ $\,b\,,$ and $\,c\,$: $$ \int_a^c f + \int_c^b f = \int_a^b f $$ [sketch illustrating the property, when $\,a \lt c \lt b\,$]

“LINEARITY OF THE INTEGRAL”
If $\,f\,$ and $\,g\,$ are integrable, then $$ \int f+g = \int f + \int g $$ “The integral of a sum is the sum of the integrals” $$ \int kf = k\int f $$ “You can slide constants out”

64.

the Definite Integral

EXAMPLE
Spse [suppose] $\,\int_0^1 f(t)\,dt = 2\,,$
$\,\int_0^4 f(t)\,dt = -6\,,$ $\,\int_3^4 f(t)\,dt = 1\,.$
Find $\,\int_1^3 f(t)\,dt\,.$
[sketch showing the relationship of the intervals
$\,[0,1]\,,$ $\,[0,4]\,,$ and $\,[3,4]\,$] $$ \int_0^4 = \int_0^1 + \int_1^3 + \int_3^4\ , $$ so $$ \int_1^3 = \int_0^4 - \int_0^1 - \int_3^4 = -6-2-1 = -9 $$ EXAMPLE
Evaluate by interpreting in terms of areas: $$ \int_{-4}^4 \sqrt{16-x^2}\ dx $$ SOLN [solution]:
Let $\,y = \sqrt{16 - x^2}\,,$ so $\,\overbrace{y^2 = 16 - x^2}^{x^2 + y^2 = 16}\,.$

Recall: $\,x^2 + y^2 = 4^2\,$ is the circle of radius $\,4\,,$
centered at $\,(0,0)\,$;
$\,y = \sqrt{16-x^2}\,$ gives the upper half of this circle. $$ \int_{-4}^4 \sqrt{16-x^2}\ dx = \frac 12(\pi\cdot 4^2) = 8\pi $$ [sketch showing the upper half of $\,x^2 + y^2 = 16\,$]

65.

the Definite Integral (precise definition)

HOW MIGHT WE ESTIMATE
AREAS BENEATH A CURVE?
[left sketch: left estimate,
points $\,(a=x_0,f(x_0))\,$ and $\,(x_3,f(x_3))\,$
on interval $\,[a,b]\,$]

[middle sketch: right estimate,
points $\,(x_1,f(x_1))\,$ and $\,(b=x_4,f(x_4))\,$
on interval $\,[a,b]\,$]

[right sketch: midpoint estimate on interval $\,[a,b]\,$]

• Cut $\,[a,b]\,$ into equal-size intervals
• Use rectangles to estimate areas on each subinterval

Left estimate: Use left-hand endpt [endpoint]
of each subinterval;
right estimate;
midpoint estimate

GENERAL NOTATION FOR
THE AREA PROBLEM
[sketch: $\,a=x_0\,,$ $\,x_1\,,$
$\,x_i\,$ (right endpt [endpoint] of $\,i^{\text{th}}\,$ subinterval),
$\,b=x_n\,$]

right estimate: $$ \begin{gather} f(x_1)\Delta x + f(x_2)\Delta x + \cdots + f(x_n)\Delta x\cr\cr \Delta x = \frac{b-a}n\ :\ \ \text{as } n\rightarrow\infty\,, \Delta x\rightarrow 0\cr x_0 = a\,,\ \ x_1 = a + \Delta x\,,\ \ x_2 = a + 2\Delta x\,,\cr \end{gather} $$ $$ \begin{align} x_n &= a + n\Delta x\cr &= a + n\bigl(\frac{b-a}n\bigr)\cr &= b \end{align} $$

66.

Summation Notation
the Definite Integral (precise definition)

SUMMATION NOTATION
... is useful for writing sums
when the terms follow a pattern

Examples: $$ \begin{gather} \underbrace{\sum_{i=1}^3}_{ \substack{\text{read: “the sum,}\cr \text{as $\,i\,$ goes}\cr \text{from $\,1\,$ to $\,3\,,$ }\cr \text{of ...} }} \kern{-15pt} (i^2 + i) = (1^2 + 1) + (2^2 + 2) + (3^2 + 3) = 20\cr\cr \sum_{i=1}^n x_i = x_1 + x_2 + \cdots + x_n\cr\cr f(x_1) + \cdots + f(x_n) = \sum_{i=1}^n f(x_i) = \sum_{j=1}^n f(x_j) \end{gather} $$ [the $\,i\,$ and $\,j\,$ in these sums are dummy variables;
conventionally, use $\,i\,,$ $\,j\,,$ $\,k\,$]

LEFT and RIGHT ESTIMATES
USING SUMMATION NOTATION
$$ \begin{gather} \text{Right:} \ \ \underbrace{f(x_1)\Delta x + \cdots + f(x_n)\Delta x}_{:= R_a^b(f,n)} = \sum_{i=1}^n f(x_i)\Delta x\cr\cr \text{Left:} \ \ \underbrace{f(x_0)\Delta x + \cdots + f(x_{n-1})\Delta x}_{:= L_a^b(f,n)} = \sum_{i=0}^{n-1} f(x_i)\Delta x \end{gather} $$ For a fct [function] $\,f\,$ that is continuous on $\,[a,b]\,,$ $$ \text{AREA} = \lim_{n\rightarrow\infty} R_a^b(f,n) = \lim_{n\rightarrow\infty} L_a^b(f,n) $$ [Note: the function names are $\,R_a^b\,$ and $\,L_a^b\,$:
they are functions of (depend on)
two variables, $\,f\,$ and $\,n\,$]

67.

the Definite Integral (precise definition)

PRECISE DEFINITION OF
A DEFINITE INTEGRAL
[sketch: $\,a = x_0\,,$ $\,x_{i-1}\,,$ $\,c_i\,,$ $\,x_i\,,$ $\,b = x_n\,,$ $\,y = f(x)\,$]

Divide $\,[a,b]\,$ into $\,n\,$ equal subintervals,
each of length $\displaystyle\,\Delta x := \frac{b-a}n\,.$

Choose $\,c_i\in[x_{i-1},x_i]\,$ (the $\,i^{\text{th}}\,$ subinterval).

Define:

$$ \substack{\text{the definite integral}\cr \text{of $\,f\,$ from $\,a\,$ to $\,b\,$}} \ =\ \int_a^b f(x)\,dx \ :=\ \lim_{n\rightarrow\infty}\ \sum_{i=1}^n f(c_i)\Delta x $$ RIEMANN SUMS
(BERNHARD $\underbrace{\text{RIEMANN}}_{\text{R}\overline{\text{EE}}\text{-MAHN}}\,,$ 1826–1866)
The sum $\ \,\displaystyle\sum_{i=1}^n f(c_i)\Delta x\,\ $ is called a Riemann sum.
A definite integral is a limit of Riemann sums!! $$ \underbrace{\underset{\downarrow}{\overset{\uparrow}{\sum}}}_{\text{stretch it out}} \quad \text{ becomes } \int \quad \substack{ \text{to remind you}\cr \text{that integration is}\cr \text{an infinite}\cr \text{summation process!}} $$

68.

the Definite Integral (precise definition)

ESTIMATING A DEFINITE INTEGRAL
Estimate $\,\displaystyle\int_0^4 \overbrace{x^2}^{f(x) = x^2}\,dx\,$ using left endpoints
for $\,n = 4\,$ approximating rectangles.

[sketch showing four approximating rectangles;
$x$-values $\,1\,,$ $\,2\,,$ $\,3\,,$ $\,4\,$;
$\,\Delta x = \frac{4-0}4 = 1\,$]

RECT # [rectangle]	$x_i$	$f(x_i)$	AREA OF RECT
1	$x_0 = 0$	$0^2 = 0$	$0\cdot 1 = 0$
2	$x_1 = 1$	$1^2 = 1$	$1\cdot 1 = 1$
3	$x_2 = 2$	$2^2 = 4$	$4\cdot 1 = 4$
4	$x_3 = 3$	$3^2 = 9$	$9\cdot 1 = 9$
			$\sum = 14$

$$ \int_0^4 x^2\,dx \cong 14 \quad \text{(an under-estimate)} $$ FINDING A DEFINITE INTEGRAL
USING THE DEFN [definition]
(FOR A VERY SIMPLE FUNCTION)
Let $$ f(x) = \begin{cases} 3\,, & x = 0\cr 0\,, & 0 \lt x \lt 2\cr 1\,, & x = 2 \end{cases} $$ $$ \begin{gather} \Delta x = \frac{2-0}n = \frac 2n\cr \text{Choose } c_i = \text{ left-hand endpts}\cr \end{gather} $$ [graph of $\,f\,$: point $\,(0,3)\,,$ zero on the interval $\,(0,2)\,,$ point $\,(2,1)\,$ ] $$ \begin{align} \int_0^2 f(x)\,dx &= \lim_{n\rightarrow\infty} \sum_{i=1}^n f(c_i)\,\Delta x\cr &= \lim_{n\rightarrow\infty} \overbrace{f\bigl(\underbrace{c_1}_{c_1 = 0}\bigr)}^{= 3}\cdot\frac 2n + \overbrace{ \underbrace{f(c_2)}_{= 0}\cdot\frac 2n + \cdots + \underbrace{f(c_n)}_{= 0}\cdot\frac 2n}^{\text{all vanish}}\cr &= \lim_{n\rightarrow\infty} 3\cdot\frac 2n\cr\cr &= 0 \end{align} $$ (Which agrees with intuition; zero area under the curve)

69.

(included for
higher-level course)

THE FUNDAMENTAL THEOREM, PART I
If $\,f'\,$ is integrable on $\,[a,b]\,,$ then $$ \begin{align} \int_a^b f'(x)\,dx &\ \ \ \ = f(b) - f(a)\cr &\overset{\text{notation}}{=} f(x)\big|_a^b \end{align} $$ [a fct [function] $\,f\,,$ whose derivative is $\,f'\,$]

Note: Integrable means
the limit of Riemann sums (card 67a) exists.

PROOF
Divide $\,[a,b]\,$ into $\,n\,$ equal subintervals, $$ a = x_0 \lt x_1 \lt \ldots \lt x_n = b\ ;\ \ \ x_i - x_{i-1} = \Delta x $$
$$ \begin{align} f(b) - f(a) &= f(x_n) \ \underbrace{- f(x_{n-1}) + f(x_{n-1}) - \cdots + f(x_1)}_{\text{add zero many times!}} - f(x_0)\cr &=\kern{-27pt}\overset{\substack{ \text{re-group}\cr \text{and multiply}\cr \text{by $\,1\,$} }}{} \left[ \frac{f(x_n)-f(x_{n-1})}{\Delta x} + \frac{f(x_{n-1}) - f(x_{n-2})}{\Delta x} + \cdots + \frac{f(x_1) - f(x_0)}{\Delta x} \right]\Delta x\cr &= \left[ \underbrace{\frac{f(x_n)-f(x_{n-1})}{x_n-x_{n-1}}}_{\text{AVROC on $\,[x_{n-1},x_n]\,$}} + \cdots + \underbrace{\frac{f(x_1) - f(x_0)}{x_1 - x_0}}_{\text{AVROC on $\,[x_0,x_1]\,$}} \right]\Delta x\cr\cr &= \ \underbrace{[f'(c_n) + \cdots + f'(c_1)]\Delta x}_{\text{Riemann sum for $\,f'\,$}}\ \ \ \ {}_{\substack{\text{for $\,c_i\in [x_{i-1},x_i]\,,$ }\cr \text{as per the Mean Value Thm}}}\cr &\rightarrow \int_a^b f'(x)\,dx\ \ \ \ \text{ as } n\rightarrow\infty \end{align} $$ [Note: AVROC = AVerage Rate Of Change;
THM = Theorem]

70.

Antiderivatives

ANTIDERIVATIVES
DEFN [definition]: A fct [function] $\,F\,$ is called
an antiderivative of $\,f\,$ on an interval $\,I\,$
iff [if and only if]
$\,F'(x) = f(x)\ \ \forall\ \ x\in I\,.$
(So ... $\,F\,$ is a function whose derivative is $\,f\,$ ... )

Note:
$\,f\,$ is an antiderivative of $\,f'\,$
$\,f+1 \underbrace{\text{is an antiderivative of $\,f'\,$}}_{\text{since the derivative of a constant is zero}}$

or ... $\,f + 7.4\,$
or ... $\,f + \text{(any constant)}\,$!!

“UNDOING” DIFFERENTIATION;
ANTIDIFFERENTIATION
Functions with the same shape have the same derivative; i.e., the derivative uniquely determines the shape of a fct [function], but not its vertical translation.

The process of going from
derivative info [information] $\,f'(x)\,$
back to function info $\,f(x) + C\,$
is called antidifferentiation.

[sketch: drop $\,y = x^2\,$ or $\,y = x^2 + 3\,$
into the ‘take the derivative’ box;
get $\,y = 2x\,$ out the bottom;
drop $\,y = 2x\,$ into the “Undo” the Derivative box,
and get $\,y = x^2 + C\,$ out the bottom]

71.

Some Basic Antidifferentiation Formulas

EVERY DIFFERENTIATION FORMULA
GIVES AN ANTIDIFFERENTIATION FORMULA!
$$ \begin{align} \frac d{dx} \sin x = \cos x\quad &\text{ so }\quad \text{$\,\sin x\,$ is a fct [function] whose derivative is $\,\cos x\,$}\cr &\text{ so } \quad \text{$\,\sin x\,$ is an antiderivative of $\,\cos x\,$}\cr &\text{ so } \quad \text{$\,F(x) = \sin x + \overbrace{C}^{\text{a constant}}\,$ is the most general antiderivative of $\,\cos x\,$} \end{align} $$ The derivative of $\,\sin x\,$ is $\,\cos x\,$;
the most general antiderivative of $\,\cos x\,$ is $\,\sin x + C\,$

SOME ANTIDERIVATIVES
YOU SHOULD KNOW

Derivative Formula	Function	Most General Antiderivative
$F'(x) = f(x)$	$f(x)$	$F(x) + C$
$\frac d{dx}\frac 1{n+1}x^{n+1} = x^n$	$x^n$	$\frac{x^{n+1}}{n+1} + C$
$\frac d{dx}\ln |x| = \frac 1x$	$\frac 1x$	$\ln |x| + C$
$\frac d{dx} \text{e}^x = \text{e}^x$	$\text{e}^x$	$\text{e}^x + C$
$\frac d{dx}\sin x = \cos x$	$\cos x$	$\sin x + C$
$\frac d{dx} (-\cos x) = \sin x$	$\sin x$	$-\cos x + C$

72.

the Definite Integral

EXAMPLES: USING THE
FUNDAMENTAL THEOREM
$$ \int_1^2 x^2\,dx = \frac{x^3}3\bigg|_1^2 = \frac 83 - \frac 13 = \frac 73 $$ [graph of $\,y = x^2\,$ showing AREA $ = \frac 73\,$
on interval $\,[1,2]\,$]
[$\frac{x^3}3\,$ is an antiderivative of $\,x^2\,$] $$ \begin{align} \int_2^5 \frac{2x^2 + 5}{x^2} &= \int_2^5 (\underbrace{2 + 5x^{-2}}_{\text{rename!}})\,dx = 2x + \frac{5x^{-1}}{-1}\bigg|_2^5\cr &= (2x - \frac 5x)\bigg|_2^5 = (2\cdot 5 - \frac 55) - (2\cdot 2 - \frac 52)\cr &= 7.5 \end{align} $$ MORE EXAMPLES
$$ \begin{align} \int_3^7 x(5x^2 + 3)\,dx &= \int_3^7 (\underbrace{5x^3 + 3x}_{\text{rename!}})\,dx\cr &= \frac{5x^4}4 + \frac{3x^2}2\bigg|_3^7 = \bigl(\frac{5\cdot 7^4}4 + \frac{3\cdot 7^2}2\bigr) - \bigl(\frac{5\cdot 3^4}4 + \frac{3\cdot 3^2}2\bigr)\cr \end{align} $$
$$ \begin{align} \int_3^6 \frac 4{\sqrt x}\,dx &= \int_3^6 4x^{-1/2}\,dx = \frac{4x^{1/2}}{1/2}\bigg|_3^6\cr &= 8x^{1/2}\bigg|_3^6 = 8[\,\sqrt{6} - \sqrt 3\,] \end{align} $$

73.

the Definite Integral

A VERY USEFUL FORMULA!!
Rewrite: $\displaystyle\,\int_a^b f'(x)\,dx = f(b) - f(a)\,$
as: $\displaystyle\,\int_a^t f'(x)\,dx = f(t) - f(a)\,$
and solve for $\,f(t)\,$: $$ f(t) = f(a) + \int_a^t f'(x)\,dx $$ If you know the derivative of a fct [function],
(so you know the shape of $\,f\,$),
and you know one point $\,(a,f(a))\,$ on the graph of $\,f\,,$
then you know all of $\,f\,$!!

INTEGRATING A RATE OF CHANGE
GIVES TOTAL CHANGE
$$ \underbrace{\int_a^b f'(x)\,dx}_{\substack{ \text{integrating a}\cr \text{rate of change}}} \ \ \ \ \underbrace{=}_{\text{gives}} \ \ \ \ \underbrace{f(b) - f(a)}_{\text{total change}} $$

So:

$$ \underbrace{f(t)}_{\substack{ \text{the value}\cr \text{of $\,f\,,$ }\cr \text{at $\,t\,$} }} = \underbrace{f(a)}_{\substack{ \text{the value}\cr \text{of $\,f\,$}\cr \text{at a known}\cr \text{place, $\,a\,$} }} + \underbrace{\int_a^t f'(x)\,dx}_{\substack{ \text{how much $\,f\,$ has changed}\cr \text{in going from $\,a\,$ to $\,t\,$} }} $$

74.

Using Antiderivatives to Find Area

CONNECTION BETWEEN
AREA AND ANTIDERIVATIVES
Let $\,f\,$ be continuous on $\,[a,b]\,$
(and assume $\,f\ge 0\,$ for this motivation).

Let

$$ \begin{align} A(x) &= \text{area under graph of $\,f\,$ on $\,[a,x]\,$}\cr &= \int_a^x f(t)\,dx \end{align} $$ [sketch: graph of $\,y = f(x)\,$ on interval $\,[a,x]\,,$
area labeled $\,A(x)\,$]

It's plausible that $\,A\,$ is an antiderivative for $\,f\,,$
since for small $\,\Delta x\,,$ $$ A'(x) \cong \frac{A(x+\Delta x) - A(x)}{\Delta x} \cong \frac{f(x)\Delta x}{\Delta x} = f(x) $$ [sketch illustrating $\,A(x+\Delta x) - A(x)\,$;
$x$-values $\,a\,,$ $\,x\,,$ $\,x+\Delta x\,,$ point $\,(x,f(x))\,$]

MAKING IT MORE PRECISE...
Since $\,f\,$ is continuous on $\,[x,x+\Delta x]\,,$
by the Extreme Value Thm [theorem]
$\,\exists\,$ [there exist] $\,x_{\text{min}}\,$ and $\,x_{\text{max}}\,$
where $\,f\,$ attains its max and min values, so $$ \underbrace{f(x_{\text{min}})\Delta x}_{\text{under-approx}} \ \le \ \underbrace{A(x+\Delta x) - A(x)}_{\text{area of thin “strip”}} \ \le \ \underbrace{f(x_{\text{max}})\Delta x}_{\text{over-approx}} $$ [Note: ‘approx’ = approximation]
[sketch: $x$-values $\,x\,,$ $\,x_{\text{max}}\,,$ $\,x+\Delta x = x_{\text{min}}\,$;
points $\,(x_{\text{max}}\,,\,f(x_{\text{max}}))\,$ and $\,(x_{\text{min}}\,,\,f(x_{\text{min}}))\,$]

So, dividing by $\,\Delta x\,,$ $$ f(x_{\text{min}}) \le \frac{A(x+\Delta x) - A(x)}{\Delta x} \le f(x_{\text{max}}) $$ As $\,\Delta x\rightarrow 0\,,$ $\,x_{\text{min}}\rightarrow x\,$ and $\,x_{\text{max}}\rightarrow x\,$;
[sketch: $\,x_{\text{min}}\,$ and $\,x_{\text{max}}\,$ are trapped in here
(between $\,x\,$ and $\,x+\Delta x\,$)]

75.

Using Antiderivatives to Find Area

MAKING IT MORE PRECISE,
CONTINUED...
Since $\,f\,$ is continuous, $$ f(x_{\text{min}})\rightarrow f(x) \quad \text{ and } \quad f(x_{\text{max}})\rightarrow f(x) $$

so

$$ f(x) \le \underbrace{\lim_{\Delta x\rightarrow 0}\frac{A(x+\Delta x) - A(x)}{\Delta x}}_{A'(x)} \le f(x) $$ so $\,A'(x) = f(x)\,.$ Thus, $\,A\,$ is an antiderivative of $\,f\,$!
If $\,F\,$ is any other antiderivative, then (since all

[second side of card]
antiderivatives differ by a constant): $$ A(x) = F(x) + C $$ Since $\,A(a) = 0\,$: $A(a) = F(a) + C\,,$ $\,C = -F(a)\,$

Then,

$$ \begin{align} A(b) &= \text{area under graph of $\,f\,$ on $\,[a,b]\,$}\cr &= F(b) + C\cr &= \kern{-8pt} \underbrace{F(b) - F(a)}_{\substack{ \text{the area depends ONLY}\cr \text{on the value of}\cr \text{ANY antiderivative}\cr \text{at the beginning and end}\cr \text{of the interval!!} }} \end{align} $$

76.

the Definite Integral

SUMMARY:
THE FUNDAMENTAL THEOREM
OF CALCULUS
... establishes a connection between
the 2 branches of Calculus

• Differential Calculus (arose from the tangent problem)
• Integral Calculus (arose from the area problem)

Let $\,f\,$ be continuous on $\,[a,b]\,.$

EXISTENCE OF AN ANTIDERIVATIVE

If $\,\displaystyle g(x) = \int_a^x f(t)\,dt\,,$
then $\,g'(x) = f(x)\,$
(so $\,g\,$ is an A.D. [AntiDerivative] of $\,f\,$)

[second side of card]

THE EVALUATION THEOREM

If $\,F\,$ is any A.D. of $\,f\,$
(i.e., $\,F'(x) = f(x)\,$ ), then $$ \int_a^b f(t)\,dt = F(b) - F(a) $$ Thus, the Fundamental Thm [theorem] has two parts:
showing that an antiderivative exists
(the “cumulative area” fct [function]
is an antiderivative),
and using any antiderivative
to evaluate a definite integral!!

77.

(included for
higher-level course)

EXAMPLES:
DERIVATIVES INVOLVING INTEGRALS

EXAMPLE.

Let $\,\displaystyle g(x) = \int_1^x \underbrace{(t^2-3)^{17}}_{:= f(t)}\,dt\,$; find $\,g'(x)\,.$

SOLN [Solution].

$\,g(x) = F(x) - F(1)\,,$ where $\,F' = f\,.$

So,

$$ \begin{align} g'(x) &= F'(x) - 0\qquad \text{(since $\,F(1)\,$ is a constant)}\cr &= f(x)\cr &= (x^2 - 3)^{17} \end{align} $$ (This approach extends to more complicated problems, as shown in 77b)

A MORE COMPLICATED DERIVATIVE
INVOLVING AN INTEGRAL
Find: $\,\displaystyle \frac d{dx} \int_{2x}^{x^2} \underbrace{\sin(5t-1)}_{:= f(t)}\,dt\,$

SOLN. Let $\,\displaystyle g(x) = \int_{2x}^{x^2} \sin(5t-1)\,dt\,.$
Then, $\,g(x) = F(x^2) - F(2x)\,,$ where $\,F' = f\,,$ so $$ \begin{align} g'(x) &= F'(x^2)\cdot 2x - F'(2x)\cdot 2\qquad\text{(chain rule)}\cr &= f(x^2)\cdot 2x - f(2x)\cdot 2\cr &= 2x\sin(5x^2-1) - 2\sin(10x - 1) \end{align} $$

78.

Antiderivatives

INDEFINITE INTEGRALS
(GENERAL ANTIDERIVATIVES)
We use the notation $\,\displaystyle\int f(x)\,dx\,$ (no limits) to denote one (or more) antiderivatives of $\,f\,$;
this is called an indefinite integral.

$\displaystyle\,\int_a^b f(x)\,dx\,$ is a number ;
$\displaystyle\,\int f(x)\,dx\,$ is a function (or class of fcts [functions])

EX [Example]:

$$ \int x^2\,dx = \overbrace{\underbrace{\frac{x^3}{3}}_{\substack{\text{one A.D.}\cr\text{[antiderivative]}}} + C\quad}^{\text{ALL the A.D.s}} $$ MORE ANTIDERIVATIVE FORMULAS
$$ \begin{gather} \int \sec^2 x\,dx = \tan x + C\cr \int \sec x\tan x\,dx = \sec x + C\cr \int \frac{1}{x^2 + 1}\,dx = \arctan x + C\cr \int\frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C \end{gather} $$

79.

Some Basic Antidifferentiation Formulas

FINDING A PARTICULAR ANTIDERIVATIVE
[graph of the line $\,y = 2x\,$] $\,f'(x) = 2x\,$
Suppose you know the derivative of a function;

[graph of several members of the class $\,y = x^2 + C\,$]
Then, $\,f(x) = x^2 + C\,$;
all these fcts [functions] have the same shape
(the same derivative).

If you are told one point on the graph of $\,f\,,$
then you can choose a particular antiderivative
from the entire class of antiderivatives.

EXAMPLE
Spse [suppose] $\displaystyle\,f'(x) = 1 + \frac{3}{x^2}\,$ and $\,f(1) = 5\,.$
Find $\,f\,.$

SOLN [solution] #1:

$$ \begin{align} f'(x) &= 1 + 3x^{-2}\cr f(x) &= x + 3\left(\frac{x^{-2+1}}{-2+1}\right) + C\cr &= x - \frac 3x + C \end{align} $$ $$f(1) = 5 \implies 1 - \frac 31 + C = 5\implies C = 7$$ So: $\displaystyle\,f(x) = x - \frac 3x + 7$

SOLN #2:

$$ \begin{align} f(x) &= f(1) + \int_1^x f'(t)\,dt = 5 + \int_1^x \bigl(1 + \frac 3{t^2}\bigr)\,dt\cr &= 5 + (t-\frac 3t)\big|_1^x = 5 + (x-\frac 3x) - (1 - 3)\cr &= x - \frac 3x + 7 \end{align} $$

80.

Some Basic Antidifferentiation Formulas

GENERALIZING THE FORMULA
FOR THE DERIVATIVE OF $\,\ln x\,$
Recall: $\,\frac d{dx}\ln x = \frac 1x\,,$ for $\,x > 0\,$ $$ \ln |x| = \cases{ \ln x\ , & \text{for } x > 0\cr \ln(-x)\ , & \text{for } x \lt 0 } $$ [sketch of graph of $\,y = \ln |x|\,$;
point at $\,x = 3\,$ with $\,m = \frac 13\,$;
point at $\,x = -3\,$ with $\,m = -\frac 13\,$]

so

$$ \frac d{dx}\ln |x| = \cases{ \frac 1x\ , & \text{for } x > 0\cr \frac 1{-x}\cdot -1 = \frac 1x\ , & \text{for } x \lt 0 } $$

So,

$$ \boxed{\ \strut\frac d{dx} \ln |x| = \frac 1x\ } $$ THE ANTIDERIVATIVES OF $\displaystyle\,\frac 1x\,$
$$ \int\frac 1x\,dx = \ln(|x|) + C $$ This formula can only be used on an interval that contains only positive, or only negative, numbers.

Thus, for example, $$ \int_{-2}^{-1}\frac 1x\,dx = \ln|x|\ \bigg|_{-2}^{-1} = \ln 1 - \ln 2 = -\ln 2\ , $$ but $\,\displaystyle\int_{-2}^2\frac 1x\,dx\,$ does not exist. Be careful!!

81.

the Substitution Technique for Integration

SUBSTITUTION
... is an integration technique.
You choose an appropriate substitution
to transform a difficult integral to an easier one.
Solve the easier problem, then transform back!

KEY IDEA: Let $\,u = \,$ something whose derivative is a factor in the integrand (maybe off by a constant).
Often, $\,u\,$ is something inside parentheses,
under a square root, in an exponent, etc.

FORMAT FOR SUBSTITUTION PROBLEMS
$$ \begin{align} \text{ORIGINAL INTEGRAL} &= \text{prepare for transformation}\cr\cr \boxed{\substack{\text{Substitution}\cr\text{Goes}\cr\text{Here}}}\qquad\quad &= \text{transform}\cr\cr &= \text{solve new problem}\cr\cr &= \text{transform back} \end{align} $$

82.

the Substitution Technique for Integration

A SUBSTITUTION EXAMPLE

$\displaystyle\overbrace{\int x(1-3x^2)^{78}\,dx}^{\text{original integral}}$	$\lower{10pt}=$	$\displaystyle\int x(1-3x^2)^{78}\cdot\overbrace{\frac{-6}{-6}}^{\text{multiply by $\,1\,$}}\,dx$
Write subst. [substitution] beneath integral: $$ \begin{gather} u = 1-3x^2\cr\cr \underbrace{\frac{du}{dx} = -6x}_{\substack{\text{not necessary}\cr \text{to write}\cr \text{this step}}}\cr\cr du = \underbrace{-6x\,dx}_{\substack{ \text{the “$\,-6\,$”}\cr \text{is missing}\cr \text{in the}\cr \text{original integral}}} \end{gather} $$	$\raise{5pt}=$	$\displaystyle -\frac 16\int (\underbrace{1-3x^2}_{u})^{78}(\underbrace{-6x\,dx}_{du})$
	$=$	$\displaystyle -\frac 16\int u^{78}\,du\quad \substack{(\text{show the}\cr \text{transformed}\cr \text{integral})}$
	$=$	$\displaystyle -\frac 16\frac{u^{79}}{79} + C$
	$=$	$\displaystyle -\frac 1{474}(1-3x^2)^{79} + C\quad \substack{ \text{(transform}\cr \text{back)}}$

SAME PROBLEM—MORE COMPACT

$\displaystyle\int x(1-3x^2)^{78}\,dx$	$\lower{10pt}=$	$\displaystyle -\frac 16\int (1-3x^2)^{78}(-6x\,dx)$
$$ \begin{gather} u = 1-3x^2\cr\cr du = -6x\,dx \end{gather} $$	$=$	$\displaystyle -\frac 16\int u^{78}\,du\,$
	$=$	$\displaystyle -\frac 16\frac{u^{79}}{79} + C$
	$=$	$\displaystyle -\frac 1{474}(1-3x^2)^{79} + C\,$

83.

the Substitution Technique Applied to Definite Integrals

SUBSTITUTION WITH DEFINITE INTEGRALS
1^st APPROACH: COMPUTE NEW LIMITS

$\displaystyle\int_2^3 t\sqrt{t-2}\,dt$	$=$	$\displaystyle\int_0^1 (u+2)\sqrt{u}\ du\quad\text{(note new limits)}$
$$ \begin{gather} u = t - 2;\ t = u+2\cr du = dt\cr\cr t = 2\,,\ u = 0\cr t = 3\,,\ u = 1 \end{gather} $$	$=$	$\displaystyle\int_0^1 (u^{3/2} + 2u^{1/2})\,du$
	$=$	$\displaystyle\frac 25u^{5/2} + 2\cdot\frac 23u^{3/2}\ \bigg|_0^1$
	$=$	$\displaystyle \frac 25 + \frac 43 = \frac 6{15}+\frac{20}{15} = \frac{26}{15}$

2^nd APPROACH: MERELY INDICATE
THAT LIMITS ARE DIFFERENT

$\displaystyle\int_0^{1/3} \frac{1}{1 + 9x^2}\,dx$	$=$	$\displaystyle \frac 13\int_0^{1/3} \frac 1{1 + (3x)^2}\,3\,dx$
$$ \begin{gather} u = 3x\cr du = 3\,dx \end{gather} $$	$\raise{27pt}=$	$\displaystyle\frac 13 \underbrace{\int_{u(0)}^{u(1/3)}}_{\substack{ \text{don't actually}\cr \text{compute}\cr \text{new limits}}} \frac 1{1+u^2}\,du = \frac 13\tan^{-1}(u)\,\bigg|_{u(0)}^{u(1/3)}$
	$\raise{27pt}=$	$\displaystyle \frac 13\,\tan^{-1}( \underbrace{3x}_{\substack{ \text{switch back}\cr \text{to original}\cr \text{variable}}} ) \, \bigg|_0^{1/3} \quad \text{(use original limits)} $
	$=$	$\displaystyle \frac 13\left[ \tan^{-1}1 - \tan^{-1}0 \right] = \frac 13(\frac\pi 4 - 0) = \frac{\pi}{12}$

84.

Integration by Parts Formula

INTEGRATION BY PARTS
... is the integration “counterpart”
to the product rule for differentiation.

Derivation:

$$ \begin{gather} \frac d{dx}(uv) = u\,\frac{dv}{dx} + v\,\frac{du}{dx}\cr\cr \text{Integrate wrt [with respect to] $x$}\cr\cr \int \frac d{dx}(uv)\,dx = \int u\,\frac{dv}{dx}\,dx + \int v\,\frac{du}{dx}\,dx\cr\cr uv = \int u\,dv + \int v\,du\cr\cr \boxed{\ \ \int u\,dv = uv - \int v\,du\ \ }\cr \text{the Integration by Parts Formula} \end{gather} $$ BASIC STRATEGY FOR USING
THE PARTS FORMULA

1. Choose something for $\,dv\,$
   that you know how to integrate
2. Choose something for $\,u\,$
   that gets simpler when you differentiate it

85.

Integration by Parts Formula

FORMAT FOR INTEGRATION BY PARTS

$\displaystyle\int \underset{\substack{ \text{‘$\,x\,$” gets simpler}\cr \text{when you differentiate it,}\cr \text{$\,\sin x\,$ doesn't!}}} { \underbrace{x}_{u} \overbrace{ \underbrace{\sin x\,dx}_{dv}}^{\substack{ \text{whatever you choose}\cr \text{for $\,dv\,$}\cr \text{must include $\,dx\,$}}} } $	$=$	$\displaystyle \underbrace{\strut x}_{u} \underbrace{(-\cos x)}_{v} - \int \underbrace{(-\cos x)}_{v} \underbrace{\strut dx}_{du} $
$$ \begin{alignat}{2} u &= x&\quad dv &= \sin x\,dx\qquad\qquad\cr du &= dx&\quad v &= \kern{-25pt}\underbrace{-\cos x}_{\substack{ \text{you can use ANY}\cr \text{antiderivative; usually,}\cr \text{you choose}\cr \text{the simplest one}\cr \text{(constant of integration $\ =\ 0\,$)} }} \end{alignat} $$	$=$	$\displaystyle -x\cos x + \int \cos x\,dx$
	$=$	$-x\cos x + \sin x + C$

USING PARTS WITH A DEFINITE INTEGRAL

$\displaystyle\int_1^{\text{e}} \overbrace{\ln x}^{u} \overbrace{dx}^{dv} $	$\kern{-5pt}=$	$\displaystyle \overbrace{(\ln x)}^{u} \overbrace{(x)}^{v}\,\bigg|_1^{\text{e}} - \int_1^{\text{e}} \overbrace{x}^{v}\ \overbrace{\frac 1x dx}^{du} $
$$ \begin{alignat}{2} u &= \ln x&\quad dv &= dx\qquad\qquad\cr\cr du &= \frac 1x\,dx&\quad v &= x \end{alignat} $$ Note: can't let $\,dv = \ln x\,dx\,,$ since we don't know how to integrate $\,\ln x\,$!	$=$	$\displaystyle (\text{e} \overbrace{\ln {\text{e}}}^{1} - 1\, \overbrace{\ln 1}^{0}) - x\,\bigg|_1^{\text{e}}$
	$=$	$\text{e} - (\text{e} - 1)$
	$=$	$1$

86.

(included for
higher-level course)

PERIODICALLY DIFFERENTIABLE FUNCTIONS
... are fcts [functions] where taking the derivative
one or more times results in
a constant multiple of the original function.

EX [example]:

$\,{\text{e}}^{3x}\,$ is a
$$ \overbrace{\text{periodically diff fct [differentiable function]}}^{pd(x)} $$ since $$\frac d{dx} {\text{e}}^{3x} = \underset{\substack{ \text{const}\cr \text{[constant]}}}{\underset{\uparrow}{3}} \cdot \underbrace{{\text{e}}^{3x}}_{\substack{ \text{orig fct}\cr \text{[original}\cr \text{function]}}} $$

EX:

$\,\sin x\,$ is $\,pd(x)\,$: $$ \begin{gather} \frac d{dx}\sin x = \cos x\,;\cr \frac d{dx}\cos x = \underset{\text{const}}{\underset{\uparrow}{-}} \overbrace{\sin x}^{\text{orig fct}} \end{gather} $$
USING PARTS WITH PRODUCTS OF
POLYNOMIALS & PERIODICALLY DIFF. FCTS [differentiable functions]
$$ \int pd(x)\, \underbrace{\text{poly}(x)}_{u}\,dx $$ [Note: Let $\,dv = pd(x)\,dx\,$]

EX:

$\displaystyle\int x^2\text{e}^x\,dx$	$=$	$\displaystyle x^2\text{e}^x - \int \text{e}^x\cdot 2x\,dx$
$$ \begin{alignat}{2} dv &= \text{e}^x\,dx &\qquad u&= x^2\cr v &= \text{e}^x& du &= 2x\,dx\quad\cr \cr dv &= \text{e}^x\,dx & u &= x\cr v &= \text{e}^x & du &= dx \end{alignat} $$	$=$	$\displaystyle x^2\text{e}^x - 2\int x\text{e}^x\,dx$
	$=$	$\displaystyle x^2\text{e}^x - 2\left[ x\text{e}^x - \int \text{e}^x\,dx \right]$
	$=$	$\displaystyle x^2\text{e}^x - 2x\text{e}^x + 2\text{e}^x + C$

87.

(included for
higher-level course)

INVERSE FUNCTIONS
WITH SIMPLER DERIVATIVES
Many inverse fcts [functions] have derivatives that are simpler than the original fcts [functions]: e.g., $$ \begin{gather} \frac d{dx}\arcsin(x) = \frac 1{\sqrt{1-x^2}}\cr \frac d{dx}\arctan(x) = \frac 1{1+x^2} \end{gather} $$ Parts [integration by parts] are often useful here: $$ \int \overbrace{\text{inverse}(x)}^u\,\overbrace{dx}^{dv} = \text{Perhaps simpler!} $$ USING PARTS FOR INVERSE FCTS

EX:

$\displaystyle \raise{18pt}{\int \overbrace{\arcsin x}^{u}\,\overbrace{dx}^{dv}}$	$\raise{2pt}=$	$\displaystyle x\arcsin(x) - \int \frac{x}{\sqrt{1-x^2}}\,dx$
PARTS: $$ \begin{alignat}{2} u &= \arcsin(x) &\quad dv &= dx\ \cr du &= \frac 1{\sqrt{1-x^2}}\,dx & v &= x \end{alignat} $$ SUBSTITUTION: $$ \begin{align} w &= 1 - x^2\cr dw &= -2x\,dx \end{align} $$	$\raise{2pt}=$	$\displaystyle x\,\arcsin(x) + \frac 12\int\frac{(-2x)}{\sqrt{1-x^2}}\,dx$
	$\raise{2pt}=$	$\displaystyle x\,\arcsin(x) + \frac 12\int w^{-1/2}\,dw$
	$=$	$\displaystyle x\arcsin(x) + \frac 12\cdot 2(1-x^2)^{1/2} + C$
	$=$	$\displaystyle x\arcsin(x) + \sqrt{1-x^2} + C$

88.

(included for
higher-level course)

USING PARTS WITH PRODUCTS
OF PERIODICALLY DIFFERENTIABLE
FCTS [functions]
Sometimes, repeated use of parts brings us back
to an expression involving the original integral,
and we can solve for it:

$\displaystyle \boxed{\ \ \int \text{e}^x\cos x\,dx\ \ }$	$=$	$\displaystyle \raise{20pt}{\text{e}^x\cos x + \overbrace{\int \text{e}^x\sin x\,dx}^{\text{use parts again}}}$
$$ \begin{alignat}{2} u &= \cos x&\quad dv &= \text{e}^x\,dx\quad \cr du &= -\sin x\,dx & v &= \text{e}^x\cr\cr u &= \sin x & dv &= \text{e}^x\,dx\cr du &= \cos x\,dx & v &= \text{e}^x \end{alignat} $$	$=$	$\displaystyle \text{e}^x\cos x + \text{e}^x\sin x - \boxed{\ \ \int \text{e}^x\cos x\,dx\ \ } $
	Solve for original integral:
	$\displaystyle \int \text{e}^x\cos x\,dx = \frac 12\text{e}^x(\cos x + \sin x) + C$

... BUT IT DOESN'T ALWAYS WORK

$\displaystyle \int \cos x\,\cos x\,dx$	$=$	$\displaystyle \cos x\,\sin x + \int \sin x\,\sin x\,dx$
$$ \begin{alignat}{2} u &= \cos x&\quad dv &= \cos x\,dx\quad \cr du &= -\sin x\,dx & v &= \sin x\cr\cr u &= \sin x & dv &= \sin x\,dx\cr du &= \cos x\,dx & v &= -\cos x \end{alignat} $$	$\displaystyle \kern{7pt}{\lower{24pt}{\underbrace{=\ \cos x\,\sin x - \sin x\cos x + \int \cos x\,\cos x\,dx}_{\text{True, but useless!!}}}} $

Instead, must use:

$$ \cos^2 x = \frac 12\bigl(1 + \cos(2x)\bigr) $$

89.

Analyzing a Falling Object

ANALYZING FALLING OBJECTS
Newton's 2^nd law says that the sum of forces acting on an object equals its mass times its acceleration: $$ \sum \text{forces} = ma $$ [sketch: a FBD (free-body diagram) showing an object with only one force acting—gravity, $\,mg\,$]

For a falling object acted on only by gravity $$ \begin{align} (g &= \text{acc [acceleration] due to gravity}\cr &\cong 9.8 \text{ m/sec}^2\cr &\cong 32.2 \text{ ft/sec}^2) \end{align} $$

where

$$ \begin{align} x(t) &= \text{position at time } t\,,\cr x(0) &= \text{original height}\,,\cr \text{positive} &\text{ direction is DOWN:} \end{align} $$ $$ \overbrace{mg = mx''(t)}^{\text{Note: mass cancels!!}} $$ (If frictional forces are neglected,
objects fall at the same rate!!)

EXAMPLE—INTEGRATING TO FIND
VELOCITY AND POSITION INFO [information]
An object is dropped from the top
of a 1000 ft bldg [building].
How long will it take to hit the ground,
and how fast will it be going when it hits?

[sketch showing building with origin at top,
+ direction is down]
$$ \begin{align} \cancel{m}g &= \cancel{m}x''(t)\,,\ \ x(0) = 0\,,\ \ x'(0) = 0\cr x''(t) &= g\cr\cr x'(t) &= gt + C\ :\ \ x'(0) = g\cdot 0 + C\,,\ \ 0 = C\cr x'(t) &= gt\ \ \text{(velocity formula)}\cr\cr x(t) &= \frac{gt^2}2 + C\ :\ \ x(0) = g\cdot 0 + C\,,\ \ 0 = C \end{align} $$

so

$$ \begin{gather} x(t) = \frac{gt^2}2\cr\cr 1000 = \frac{(32.2)t^2}2\cr\cr \implies t \cong 7.9 \text{ sec}\cr\cr x'(7.9) = g(7.9) \cong 254 \text{ ft/sec} \end{gather} $$

Index Card Titles/Concepts

In each entry, the first line is side (a), the second line is side (b). Two cards (43 and 53) have additional sides (c) and (d).

1.

average rate of change;
the tangent problem

2.

position functions;
the derivative of a function

3.

intuitive derivative (notation);
example

4.

limit of a function;
3 cases when $\,\lim_{x\rightarrow a} f(x) = \ell\,$

5.

one-sided limits;
the squeeze (pinching) theorem

6.

continuity at a point;
three ways a function can fail to be continuous at $\,c\,$

7.

key idea (for continuity at a point);
example

8.

an important type of continuity problem;
relationship between differentiability and continuity

9.

limit rules;
techniques for evaluating limits (examples)

10.

indeterminate forms;
deciding if you have an ‘indeterminate form’

11.

the derivative of $\,f\,$ at $\,a\,$;
using the definition of derivative

12.

an alternative approach (compare with 11b);
another example: using the definition of derivative

13.

the linearization of $\,f\,$ at $\,a\,$;
intuition for the linearization

14.

local max/min (extreme values);
global max/min

15.

strictly increasing/decreasing functions;
getting increasing/decreasing behavior from the derivative

16.

concavity;
getting concavity info from the second derivative

17.

power rule for differentiation;
examples: using the power rule

18.

partial proof of the power rule (both sides)

19.

derivative of a constant times a function;
finding $\,\frac{d}{dx} (kx^n)\,$

20.

derivatives of sums and differences;
proof

21.

differentiating composite functions: motivation;
differentiating composite functions: the idea

22.

the chain rule (prime notation);
the chain rule (Leibnitz notation)

23.

Why is it called the ‘chain rule’?
Using the chain rule to differentiate $\,\bigl( f(x) \bigr)^n\,$

24.

pattern for generalizing all the basic differentiation formulas;
examples

25.

the product rule for differentiation;
proof of the product rule

26.

When is $\,\lim_{h\rightarrow 0} f(x+h) = f(x)\,$?
When is this true? As $\,x\rightarrow c\,,$ $\,f(x)\rightarrow f(c)\,$

27.

the quotient rule for differentiation;
proof of the quotient rule

28.

the irrational number $\,\text{e}\,$;
equivalent limit statements defining $\,\text{e}\,$

29.

evaluating a limit involving $\,\text{e}\,$;
Why is $\,\lim_{n\rightarrow\infty} \left[ (1 + \frac 1n)^n \right]^{2k} = \text{e}^{2k}\,$?

30.

derivatives of logarithms;
proof that $\,\frac d{dx} (\ln x) = \frac 1x\,$

31.

more notation for derivatives;
notation: evaluating a derivative at a point

32.

differentiating $\,{\text{e}}^x\,$;
differentiating $\,a^x\,$

33.

what is $\,\lim_{x\rightarrow 0} \frac{\sin x}{x}\,$?
example

34.

derivatives of sine and cosine;
derivatives of tangent and cotangent

35.

derivatives of secant and cosecant
a useful observation—derivatives of co-functions

36.

finding the derivative of an inverse function;
two methods: finding the derivative of an inverse function

37.

What does $\,(f^{-1})'(x) = \frac{1}{f'\bigl(f^{-1}(x)\bigr)}\,$ really mean?
summary

38.

derivatives of the inverse trigonometric functions;
a typical derivation

39.

differentiating variable stuff to variable powers: the log trick;
example

40.

l'Hopital's rule;
motivation, ‘$\,\frac 00\,$’ case

41.

basic use of l'Hopital's rule;
more advanced use of l'Hopital's rule

42.

Where can a function change its sign?
negating ‘and’ and ‘or’ sentences

43ab.

(a) info given by $\,f\,,$ $\,f'\,,$ $\,f''\,$;
(b) a basic sign analysis of a function

43cd.

(c) implications (and equivalent sentences);
(d) the contrapositive of an implication

44.

Where can a function have a local max/min?
If $\,f\,$ has a local max/min at $\,c\,,$ then ...

45.

critical points, critical numbers;
Careful!! A critical point does not have to be a max or min!!

46.

Where can a function have an inflection point?
If $\,f\,$ has an inflection point at $\,c\,,$ then $\,\bigl(f''(c) = 0\ \ \text{or}\ \ f''(c) \text{ DNE}\bigr)\,.$

47.

an algorithm for thorough function analysis;
example—thorough function analysis

48.

Are critical points actually max or min? Two tests: 1st and 2nd derivative tests;
the first derivative test

49.

open/closed intervals;
bounded subsets of $\,\Bbb R\,$

50.

review of absolute/global max/min;
the Extreme Value Theorem

51.

optimization problems (finding max/min);
example

52.

a first optimization problem;
You try it!

53ab.

(a) an optimization problem;
(b) ... continued

53cd.

(c) a useful observation;
(d) using the observation

54.

an optimization problem;
... continued

55.

explicit (‘showing’) versus implicit (‘hidden’);
implicit differentiation

56.

related rate problems;
falling ladder problem

57.

Is the car speeding?
... continued

58.

the Mean Value Theorem;
the Mean Value Theorem (MVT) intuition

59.

a Mean Value Theorem example;
a Mean Value Theorem example

60.

bounded functions;
the definite integral of a function

61.

a definite integral may or may not exist;
notation for the definite integral

62.

dummy variables in definite integrals;
simple definite integral examples

63.

properties of the definite integral;
linearity of the integral

64.

definite integral example;
definite integral example

65.

How might we estimate areas beneath a curve?
general notation for the area problem

66.

summation notation;
left and right estimates using summation notation

67.

precise definition of a definite integral;
Riemann sums

68.

estimating a definite integral;
finding a definition integral using the definition (for a very simple function)

69.

the Fundamental Theorem, Part I;
proof

70.

antiderivatives;
‘undoing’ differentiation; antidifferentiation

71.

Every differentiation formula gives an antidifferentiation formula!
some antiderivatives you should know

72.

examples: using the Fundamental Theorem;
more examples

73.

A very useful formula!!
integrating a rate of change gives total change

74.

connection between area and antiderivatives;
making it more precise...

75.

making it more precise, continued (both sides)

76.

summary: the Fundamental Theorem of Calculus, Existence of an Antiderivative;
the Evaluation Theorem

77.

examples: derivatives involving integrals;
a more complicated derivative involving an integral

78.

indefinite integrals (general antiderivatives);
more antiderivative formulas

79.

finding a particular antiderivative;
example

80.

generalizing the formula for the derivative of $\,\ln x\,$;
the antiderivatives of $\,\frac 1x\,$

81.

substitution;
format for substitution problems

82.

a substitution example;
same problem, more compact

83.

substitution with definite integrals;
2nd approach: merely indicate that limits are different

84.

integration by parts;
basic strategy for using the parts formula

85.

format for integration by parts;
using parts with a definite integral

86.

periodically differentiable functions;
using parts with products of polynomials and periodically differentiable functions

87.

inverse functions with simpler derivatives;
using parts for inverse functions

88.

using parts with products of periodically differentiable functions;
... but it doesn't always work

89.

analyzing falling objects;
example: integrating to find velocity and position information