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341
documents/by-course/math-6a/course-notes/dvd.typ
View file

@ -1,341 +0,0 @@
#import "@preview/ctheorems:1.1.3": *
#import "@preview/showybox:2.0.3": showybox
#let colors = (
rgb("#9E9E9E"),
rgb("#F44336"),
rgb("#E91E63"),
rgb("#9C27B0"),
rgb("#673AB7"),
rgb("#3F51B5"),
rgb("#2196F3"),
rgb("#03A9F4"),
rgb("#00BCD4"),
rgb("#009688"),
rgb("#4CAF50"),
rgb("#8BC34A"),
rgb("#CDDC39"),
rgb("#FFEB3B"),
rgb("#FFC107"),
rgb("#FF9800"),
rgb("#FF5722"),
rgb("#795548"),
rgb("#9E9E9E"),
)
#let dvdtyp(
title: "",
subtitle: "",
author: "",
abstract: none,
bibliography: none,
paper-size: "a4",
date: "today",
body,
) = {
set document(title: title, author: author)
set std.bibliography(style: "springer-mathphys", title: [References])
show: thmrules
set page(
numbering: "1",
number-align: center,
header: locate(loc => {
if loc.page() == 1 {
return
}
box(stroke: (bottom: 0.7pt), inset: 0.4em)[#text(
font: "New Computer Modern",
)[
*#author* --- #datetime.today().display("[day] [month repr:long] [year]")
#h(1fr)
*#title*
]]
}),
paper: paper-size,
// The margins depend on the paper size.
margin: (
left: (86pt / 216mm) * 100%,
right: (86pt / 216mm) * 100%,
),
)
set heading(numbering: "1.")
show heading: it => {
set text(font: "Libertinus Serif")
block[
#if it.numbering != none {
text(rgb("#2196F3"), weight: 500)[#sym.section]
text(rgb("#2196F3"))[#counter(heading).display() ]
}
#it.body
#v(0.5em)
]
}
set text(font: "New Computer Modern", lang: "en")
show math.equation: set text(weight: 400)
// Title row.
align(center)[
#set text(font: "Libertinus Serif")
#block(text(weight: 700, 26pt, title))
#if subtitle != none [#text(12pt, weight: 500)[#(
subtitle
)]]
#if author != none [#text(16pt)[#smallcaps(author)]]
#v(1.2em, weak: true)
#if date == "today" {
datetime.today().display("[day] [month repr:long] [year]")
} else {
date
}
]
if abstract != none [
#v(2.2em)
#set text(font: "Libertinus Serif")
#pad(x: 14%, abstract)
#v(1em)
]
set outline(fill: repeat[~.], indent: 1em)
show outline: set heading(numbering: none)
show outline: set par(first-line-indent: 0em)
show outline.entry.where(level: 1): it => {
text(font: "Libertinus Serif", rgb("#2196F3"))[#strong[#it]]
}
show outline.entry: it => {
h(1em)
text(font: "Libertinus Serif", rgb("#2196F3"))[#it]
}
// Main body.
set par(
justify: true,
spacing: 0.65em,
first-line-indent: 2em,
)
body
// Display the bibliography, if any is given.
if bibliography != none {
show std.bibliography: set text(footnote-size)
show std.bibliography: set block(above: 11pt)
show std.bibliography: pad.with(x: 0.5pt)
bibliography
}
}
#let thmtitle(t, color: rgb("#000000")) = {
return text(
font: "Libertinus Serif",
weight: "semibold",
fill: color,
)[#t]
}
#let thmname(t, color: rgb("#000000")) = {
return text(font: "Libertinus Serif", fill: color)[(#t)]
}
#let thmtext(t, color: rgb("#000000")) = {
let a = t.children
if (a.at(0) == [ ]) {
a.remove(0)
}
t = a.join()
return text(font: "New Computer Modern", fill: color)[#t]
}
#let thmbase(
identifier,
head,
..blockargs,
supplement: auto,
padding: (top: 0.5em, bottom: 0.5em),
namefmt: x => [(#x)],
titlefmt: strong,
bodyfmt: x => x,
separator: [. \ ],
base: "heading",
base_level: none,
) = {
if supplement == auto {
supplement = head
}
let boxfmt(name, number, body, title: auto, ..blockargs_individual) = {
if not name == none {
name = [ #namefmt(name)]
} else {
name = []
}
if title == auto {
title = head
}
if not number == none {
title += " " + number
}
title = titlefmt(title)
body = [#pad(top: 2pt, bodyfmt(body))]
pad(
..padding,
showybox(
width: 100%,
radius: 0.3em,
breakable: true,
padding: (top: 0em, bottom: 0em),
..blockargs.named(),
..blockargs_individual.named(),
[
#title#name#titlefmt(separator)#body
],
),
)
}
let auxthmenv = thmenv(
identifier,
base,
base_level,
boxfmt,
).with(supplement: supplement)
return auxthmenv.with(numbering: "1.1")
}
#let styled-thmbase = thmbase.with(
titlefmt: thmtitle,
namefmt: thmname,
bodyfmt: thmtext,
)
#let builder-thmbox(color: rgb("#000000"), ..builderargs) = styled-thmbase.with(
titlefmt: thmtitle.with(color: color.darken(30%)),
bodyfmt: thmtext.with(color: color.darken(70%)),
namefmt: thmname.with(color: color.darken(30%)),
frame: (
body-color: color.lighten(92%),
border-color: color.darken(10%),
thickness: 1.5pt,
inset: 1.2em,
radius: 0.3em,
),
..builderargs,
)
#let builder-thmline(
color: rgb("#000000"),
..builderargs,
) = styled-thmbase.with(
titlefmt: thmtitle.with(color: color.darken(30%)),
bodyfmt: thmtext.with(color: color.darken(70%)),
namefmt: thmname.with(color: color.darken(30%)),
frame: (
body-color: color.lighten(92%),
border-color: color.darken(10%),
thickness: (left: 2pt),
inset: 1.2em,
radius: 0em,
),
..builderargs,
)
#let problem-style = builder-thmbox(
color: colors.at(11),
shadow: (offset: (x: 2pt, y: 2pt), color: luma(70%)),
)
#let exercise = problem-style("item", "Exercise")
#let problem = exercise
#let theorem-style = builder-thmbox(
color: colors.at(6),
shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
)
#let example-style = builder-thmbox(
color: colors.at(16),
shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
)
#let theorem = theorem-style("item", "Theorem")
#let lemma = theorem-style("item", "Lemma")
#let corollary = theorem-style("item", "Corollary")
#let definition-style = builder-thmline(color: colors.at(8))
// #let definition = definition-style("definition", "Definition")
#let proposition = definition-style("item", "Proposition")
#let remark = definition-style("item", "Remark")
#let observation = definition-style("item", "Observation")
// #let example-style = builder-thmline(color: colors.at(16))
#let example = example-style("item", "Example")
#let proof(body, name: none) = {
v(0.5em)
[_Proof_]
if name != none {
[ #thmname[#name]]
}
[.]
body
h(1fr)
// Add a word-joiner so that the proof square and the last word before the
// 1fr spacing are kept together.
sym.wj
// Add a non-breaking space to ensure a minimum amount of space between the
// text and the proof square.
sym.space.nobreak
$square.stroked$
v(0.5em)
}
#let fact = thmplain(
"item",
"Fact",
titlefmt: content => [*#content.*],
namefmt: content => [_(#content)._],
separator: [],
inset: 0pt,
padding: (bottom: 0.5em, top: 0.5em),
)
#let abuse = thmplain(
"item",
"Abuse of Notation",
titlefmt: content => [*#content.*],
namefmt: content => [_(#content)._],
separator: [],
inset: 0pt,
padding: (bottom: 0.5em, top: 0.5em),
)
#let definition = thmplain(
"item",
"Definition",
titlefmt: content => [*#content.*],
namefmt: content => [_(#content)._],
separator: [],
inset: 0pt,
padding: (bottom: 0.5em, top: 0.5em),
)

505
documents/by-course/math-6a/course-notes/main.typ
View file

@ -98,9 +98,9 @@ $
kappa (t) = (||arrow(c)'(t) times arrow(c)''(t)||) / (||arrow(c)'(t)||^3)
$
== Arclength parameterization
== Arclength parametrization
Find an arc-length parameterization of $arrow(c)(t) = <e^t sin(t), e^t cos(t), 5e^t>$.
Find an arc-length parametrization of $arrow(c)(t) = <e^t sin(t), e^t cos(t), 5e^t>$.
Let $s = 0$ when $t = 0$ and let $s$ be the arc-length that has traveled along
the curve after $t$ seconds, then we can find $s$ by integrating the curve's
@ -273,3 +273,504 @@ the directional derivative is zero.
critical points into $f$ and find the biggest/smallest.
]
= Midterm 2 review
I literally forgot everything...time to cram.
== Vector review
We know about functions $y = f(x)$. We can parameterize functions by expressing
them as a pair of coordinates $(x(t), y(t))$, modeling for example a particle
traveling through space with respect to time.
=== Derivative of parameterized curve
Given $(x(t), y(t))$, the derivative is given
$
(x'(t), y'(t))
$
=== Parameterizing an ellipse
Consider an ellipse
$
x^2 / n + y^2 / m = r^2
$
Then we note that this is just a circle with $x$ stretched by a factor of
$sqrt(n)$ and likewise for $y$ by a factor of $sqrt(m)$. Then we can
parameterize the ellipse by
$
(r sqrt(n) cos(theta), r sqrt(m) sin(theta))
$
A sanity check: when $n = 1$, $m = 1$, $r = 1$, we have a unit circle and the
parametrization reflects that.
#example[
Consider the ellipse $x^2/9 + y^2/4 = 16$. Then the parametrization is
$(12cos(theta), 8sin(theta))$, and this is indeed right.
]
To parameterize a line passing through points $arrow(p)_1$ and $arrow(p)_2$,
simply
$
arrow(c) (t) = arrow(p)_1 + t(arrow(p)_2 - arrow(p)_1)
$
Algebraically we can justify this by noting $t=0$ gives $arrow(p)_1$ and $t=1$ gives $arrow(p)_2$.
=== Polar coordinates
Notation: $(r,theta)$ instead of $(x,y)$. Note that this is just highlighting
that we're parameterizing a curve in terms of a radius $r$ (also called
_modulus_)and argument (angle) $theta$.
To get $r$, see that $r^2 = x^2 + y^2$ and it follows that $theta =
arctan(y/x)$. Blah blah.
To plot in terms of $x,y$, note that
$
x = r cos(theta) \
y = r sin(theta)
$
=== Vector properties
Many nonrigorous statements about vectors to add to our toolbox.
Two vectors are parallel if and only if the angle formed between them is 0.
Vectors can be added in linear combinations.
The magnitude of a vector length $n$ is given by the $n$ dimensional
Pythagorean theorem.
$
sqrt(a_1^2 + a_2^2 + dots.c + a_n^2)
$
A unit vector is a vector with magnitude 1.
=== Dot product
Dot products are useful for seeing properties about orthogonality, parallelism,
and projection.
#definition[
The dot product takes two vectors and returns a single scalar. It is sometimes
called the inner product.
]
We can view the dot product algebraically, and geometrically. In the first
sense, it's just the sum of the products of each pair of coordinates in the
vectors.
Let $A,B$ be vectors length $n$, and $a_i, b_i$ be the $i^"th"$ entry of their
respective vectors, then
$
A dot B = a_1 dot b_1 + a_2 dot b_2 + dots.c + a_n dot b_n
$
Geometrically, it's the product of the magnitudes of the vectors and the cosine
of angle between them.
$
A dot B = |A| dot |B| dot cos(theta)
$
where $theta$ is the angle between the vectors. It's nontrivial to prove this
and we don't have time.
Therefore, we know two vectors are parallel when $A dot B = |A| dot |B|$,
because $cos(theta) = 1$. We know two vectors are orthogonal if the dot product
is 0, because $cos(pi/2 + pi n)$ is 1.
Dot products are very useful for projection. Pick a normal vector $arrow(u)$.
For any vector $arrow(w)$, $arrow(u) dot arrow(w)$ gives the size of the
parallel part.
Also, note this gives the shortest distance between the tip of $arrow(w)$ and
the line passing in the direction of $arrow(u)$!
=== Normalizing vectors
Normalizing a vector means obtaining a vector pointing in the same direction,
with magnitude 1. For a nonzero vector $v$, its normalized vector is given
$
v / (|v|)
$
where $|v|$ is the magnitude.
=== Cross product
The cross product $times$ is a binary operation on vectors. For our purposes
it's only defined in $RR^3$ (in fact it's defined in some other dimensions, but
in general it is not defined).
It produces a third vector perpendicular to both original vectors. Its
direction is determined by the right hand rule.
The magnitude of the cross product is given by
$
|a times b| = |a| |b| sin theta
$
where $theta$ is the angle between $a$ and $b$. So we know two vectors are
parallel if the magnitude of the cross product is 0.
In fact this magnitude is also the area of the parallelegram spanned by the two
vectors. This is an alternative way to view the cross product being 0 implying
the vectors parallel.
We can compute the cross product by taking a determinant.
$
a times b = det mat(i,j,k; a_1, a_2, a_3; b_1,b_2,b_3)
$
The cross product is anticommutative, so $a times b = -(b times a)$.
== Vector applications, building geometric intuition
Let's build some geometric intuition for working with vectors, especially in
$RR^3$.
=== Moving the equation of a line or plane while maintaining orientation
We consider two cases. If we're working with a parametric equation, then we can
just add a vector to the equation to shift everything by said vector.
Otherwise, with an implicit equation, let's consider only the plane (since we
need two implicit equations to specify a line, and at that point it's better to
solve the system and parameterize).
The plane equation $a x + b y + c z = r$ can be shifted $n$ units in the
positive $x$, $y$, or $z$ direction by replacing all $x$, $y$, or $z$ with $x -
n$ and so on. Then we can just multiply out collect terms.
=== Moving equation of a line/plane to pass through a specific point
We want to do this without changing direction or orientation. We can just use
our technique discussed above for this.
First let's make sure the plane passes through the origin. If we have
$
3x - 2y + 7z = 12
$
we can just set the right hand to $0$
$
3x - 2y + 7z = 0
$
Note that by our technique of shifting the plane or line, we see that the
constant on the right side is determined entirely by shifts in space that
preserve direction/orientation. So we are sure that setting it to 0 does
nothing but move the plane/line through 0.
=== Equation of a line through a given point perpendicular to a
plane
Let's say we have a point $(5,2,3)$. Let's consider both parametric planes and
implicitly defined planes.
Suppose the plane is given by
$
2x + 3y + 4z = 12
$
Then observe that we need the line to be perpendicular to the plane but it
doesn't really matter where the plane is. Recall that we can easily shift a
plane around while preserving orientation. So let's just move the plane through
the origin again.
$
2x + 3y + 4z = 0
$
Now note that we can obtain a perpendicular vector to the plane by finding a
vector perpendicular to any particular vector on this plane.
Then note that $vec(2,3,4)$ is one such perpendicular vector. See this by
$
vec(2,3,4) dot vec(x,y,z) = 2x + 3y + 4z = 0
$
Then we can simply scale our perpendicular vector by a parameter $t$ to obtain
a parametric line that's perpendicular to the plane. Now we can just shift it
by our desired point, and it remains orthogonal while passing through the line
(at $t = 0$).
$
vec(5,2,3) + t vec(2,3,4)
$
Now consider when we have a parametric equation, say
$
vec(5,0,0) + s vec(2,0,-1) + t vec(0,4,-3)
$
Then as long as we're perpendicular to both of the vectors being multiplied by
$s$ and $t$, we're perpendicular. This is easy to show, just note that the
plane is given by the span of the basis vectors $vec(2,0,-1)$ (shifted by
$vec(5,0,0)$) and $vec(0,4,-3)$, so any vector perpendicular to both is
perpendicular to the entire plane.
So just take their cross product to get a desired vector.
=== Distance between point and a plane
Think geometrically. We really want to move in a perpendicular line from the
plane to the point (because that's the closest distance between them). We
should start on the point, but where do we stop on the plane?
Consider
$
4x + y + 3z = 1
$
and we want the distance to $(1,1,-5)$. The perpendicular line passing through the point is
$
vec(1,1,-5) + t vec(4,1,3)
$
The line "starts" at the point $t=0$, so let's find a value of $t$ that makes
it stop precisely on the plane. To do this, simply note that our line is really a parametric equation
$(x,y,z)$ where
$x = 1 + 4t, y = 1 + t, z = -5 + 3t$. Then we can simply plug these into the
equation of the plane and solve for $t$ to get the value of $t$ where the line
meets the plane. Then plug $t$ into our line equation (which gives a vector)
and the magnitude is the distance between the point and plane.
=== Area of a parallelogram formed by two vectors in $RR^3$
In $RR^2$ we can take the determinant. In $RR^3$ the determinant is the volume
of the parallelepiped. So instead we just take the magnitude of the cross
product.
=== Distance from point to line passing through two other points in $RR^2$
Note that there are multiple ways to do this. Let $P = (1,7)$, $A = (1,1)$, and
$B = (3,9)$. We want the distance from $P$ to the line between $A$ and $B$.
We could just find the line between them and then use a 2-dimensional version
of our point to plane technique (solve for a vector orthogonal to the line, in
the direction of $P$, passing through $P$), but since we're in $RR^2$, we can
just project $P$ onto the normalized line and do some stuff.
In particular, note that the magnitude of the cross product of $A times B$ is
$|A| dot |B| dot sin theta$. So if we want the distance from the tip of $B$ to
the line spanned by $A$, we should do $(|A times B|)/(|A|)$.
If instead we want the length of the projection of $B$ onto $A$, we should do
$(A dot B)/(|A|)$. There are multiple ways to interpret this geometrically.
== Derivative of a curve
What is the derivative of a curve? We can view the derivative at some point $x$
as the slope of the tangent line. But that doesn't give the derivative of a
parametric curve traveling through the plane.
However, this is simple. Because our curve is parameterized, each coordinate
$x,y,z$ and so on is independent of each other and given by $t$. Therefore, we
can collect another vector, taking the derivative of each coordinate, which
gives us a vector of the rates at which each coordinate is changing.
== Arclength parametrization
We want to find an arc length parametrization of a curve. That is, we want to
express a curve in terms of how far we've traveled on it.
Idea: let $s=0$ when $t=0$, and let $s$ be the arclength traveled after $t$
seconds. Then we can integrate the curve's speed over $t$ to find the arc
length.
$
s(t) = integral_0^t ||arrow(c)'(u)|| dif u
$
Then, we can solve for $t$ in terms of $s$, and plug it back into our original
vector in terms of $t$, $arrow(c)(t)$. Then its position will be expressed by
in terms of $s$, $arrow(c)(s)$, and we'll have a parametrization by arc
length.
A key notion here is now the velocity vectors are tangent, but also unit length
(since we should imagine that we are always moving at unit speed along the
curve at any given point).
What direction does the _acceleration_ vector point in, then?
For a parameterized curve $arrow(c)(t)$ with velocity $arrow(v)(t)$ and
acceleration $arrow(a)(t)$, then the speed is magnitude $|arrow(v)(t)|$. When
the speed is constant, $|arrow(v)(t)|$ doesn't change with time.
A prototypical example: consider uniform circular motion. Then the angular
velocity (speed) is always constant, yet there is always an acceleration vector
pointing perpendicular to the tangent velocity vector (the centripetal
acceleration).
== Curvature
The curvature in $RR^2$ is given by the second derivative. But this is just a
lucky coincidence. Let's think about the notion of curvature.
Somehow, the curvature measures the best-fitting second order approximation of
a curve. Curvature is a measure of concavity with respect to the direction
perpendicular to the direction of motion.
We do have a formula
$
kappa(t) = (|arrow(c)'(t) times arrow(c)'' (t)|) / (|arrow(c)'(t)|^3)
$
== Building intuition for curvature
Curvature is essentially asking how closely our curve resembles a unit circle
at a given point. It follows that a unit circle has a curvature 1, and a
straight line has curvature 0.
Let's consider a parametric curve
$
arrow(s)(t) = vec(t-sin(t), 1-cos(t))
$
Consider the unit tangent vectors to the curve at some points. We're
essentially asking "how much do these tangent vectors change direction?" and
considering points arbitrarily close. This is the essence of curvature.
Now let's think about some geometric intuition. Suppose you're moving along a
curve. It follows that if the curve is bending very sharply, the tangent
vectors are changing direction very fast, in larger increments. Likewise, if
the curve is straighter, the tangent vectors change direction less often. And
when you're traveling on a straight line, the tangent vectors don't change
direction at all.
We want a mathematical model of this notion of "changing tangent vectors." The
idea is that we can capture this with some sort of derivative, but with respect
to what? If we just want to capture when the tangent vector is changing
directions, we clearly want to ignore any change in the actual magnitude of the
tangent vector itself, since this has no bearing on the directional change. Put
another way, if you're traveling along the curved path, the speed at which you
go (the magnitude of the tangent velocity vector) really doesn't matter with
regard to curvature. We only care when the tangent velocity vector changes
direction!
So suppose $T$ is the unit tangent vector at each point. We want the rate of
change of $T$, its derivative, but _not_ $(dif T)/(dif t)$, with respect to
time. This is because we don't really care how _fast_ the tangent vector
changes with respect to time, curvature is about measuring how much the tangent
vector changes as we move some arbitrary distance on the curve!
Instead, we really want $(dif T)/(dif s)$, where $s$ is the arc length we've
traveled so far (from some arbitrarily chosen starting point). And this makes
intuitive sense, because we just want how fast the tangent vector changes
direction with regards to the distance we travel on the curve.
Now we may note that we can actually find curvature if we can find an
arc-length parametrization of $arrow(s)$! Because an arclength
parametrization always has unit speed, its derivative gives the tangent
vectors at every point, and we can differentiate with respect to arclength and
take the magnitude to obtain the curvature. That is,
$
kappa = abs((dif T) / (dif s))
$
But if we can't find an arclength parametrization, we're out of luck. Let's
continue investigating.
Consider a prototypical example:
#example[
Let $arrow(s)(t) = vec(cos(t) R, sin(t) R)$. We're drawing a circle with
radius $R$.
Let's differentiate with respect to $t$.
$
arrow(s)'(t) = vec(-sin(t) R, cos(t) R)
$
But we want unit tangent vectors, so let's normalize it. Call our unit
tangent $T$.
$
T(t) = (arrow(s)'(t)) / (|arrow(s)'(t)|)
$
We have
$
|arrow(s)'(t)| = lr(|vec(-sin(t) R, cos(t) R)|) \
= sqrt(sin^2(t) R^2 + cos^2(t) R^2) = R
$
Now we have our unit tangent vectors in terms of $t$.
$
T(t) = (arrow(s)'(t)) / R
$
We take
$
(dif T) / (dif t) = vec(-cos(t), -sin(t))
$
and the magnitude of this is just 1.
We should immediately note that not all cases will be so easy. When taking
$|arrow(s)'(t)|$, in general, we have a very disgusting square root that cannot
be simplified.
Now note:
$
abs((dif T)/(dif s)) = abs((dif T)/(dif t)) / abs((dif T)/(dif s))
$
So in fact $kappa = 1/R$!
]
The key here is this equation:
$
abs((dif T)/(dif s)) = abs((dif T)/(dif t)) / abs((dif s)/(dif t))
$
Although we didn't have an arclength parametrization of $T$, we note that its
magnitude is essentially given by the magnitude at which it's changing with
respect to time, and divided by the rate the curve is moving to "correct" for
the discrepancies introduced by taking the derivative with respect to time!
Obviously this is very nonrigorous. But I'm running out of time.
Now if we do a bunch more reasoning and nonsense we can obtain the formula
above, but at this point the goal seems to have been reached. We have an
intuitive understanding of what curvature should measure.
So recall the formula:
$
kappa(t) = (|arrow(c)'(t) times arrow(c)'' (t)|) / (|arrow(c)'(t)|^3)
$
Essentially, it's saying that the area of the parallelogram formed by the
curve's velocity and acceleration vectors, divided by the cube of the speed,
gives us the curvature. Intuitively we see that the more the acceleration
vector diverges from the velocity vector, the sharper the velocity vector is
changing direction, which gives us a notion of curvature. And somehow dividing
by the speed cubed is normalizing out any influence due to speed to give us our
curvature.
== Quadric surfaces
We really only need to know the identities and derivatives to do some integral
hacks.
The quadric surface is the generalization of the conic section to $n$ dimensions.
Now recall that one conic section is the hyperbola. It turns out we can define
analogues of the trigonometric functions that parameterize a so-called unit
hyperbola instead of the unit circle. These functions are
$
sinh(x) = (e^x -e^(-x)) / 2 \
cos(x) = (e^x + e^(-x)) / 2 \
tanh(x) = sinh(x) / cosh(x)
$
The derivatives are
$
(dif) / (dif x) sinh(x) = cosh(x) \
(dif) / (dif x) cosh(x) = sinh(x)
$
It's pretty easy to show these using their definitions, and derive the
derivative of $tanh$.