Geology: Calculus Video #4

3Blue1Brown Intro To Calculus Series Summary

Carlos Manuel Jarquín Sánchez
6 min readAug 24, 2023

Preface ✨

Hello everyone! My intentions for writing these articles are:

  • Explain technical knowledge about geology in simple terms (to the public)
  • Document my journey to getting an AD-T Geology degree in one year
  • Store information and habits for my future self and others (in <7 minutes)

Coolio? Sweet. Enjoy the series :-)

notes as i watched the video 📹

p.s. — the video will be linked at the bottom of the article. cheers.

and yes, derivatives like the ones i wrote in the last article, they’re not meant to be so trivial & easy.

in the real world, they’re quite a migraine lol

so what can we use to simplify & solve the complex derivatives?

we can combine functions.

and there’s three preferred ways to get there:

  • adding functions
  • multiplying functions
  • compose functions (aka throw a function inside of another function)

here’s some examples for each one:

  • add: d/dx (sin(x) + x²) ||| derivative will be: (cos(x) + 2x)
  • multiply: d/dx (sin(x) * x²) ||| derivative will be: (sin(x) * 2x) + (cos(x) * x²)
  • compose: d/dx (sin(x²)) ||| derivative will be: cos(x²) * 2x

yes, i know i’m not explaining what i did to get from point a to point b.

i’ll give y’all my two cents in a sec.

just wanted to say that everything comes in layers, where one must peel it all back to see what the core is made out of.

like this for example:

d/dx ((e^sin(x)) * cos((1/x³) + x³)))⁴

it’s a big mouthful at first, but if we break it down more…

  • the outermost bracket, ()⁴, is a composition
  • e^sin(x) (e to the power of sin(x)), is multiplying functions
  • the outermost bracket for cosine, cos (), is a composition
  • “+ x³” is adding a function
  • 1/x³, or 1*(1/3), is a composition (combination) of a variable & fraction

…it’s not so intimidating.

we just gotta know what layer is what.

let’s start with adding derivatives (or miniscule changes in value) together.

f(x) = sin(x) + x²

“x” will have a value of 1.5

let’s say we add a bit more to “x”, now it will no longer be 1.5

it will be 1.5 PLUS whatever extra we added onto it, but because we don’t know how much we added on, we use “dx”.

so u can say it went from:

f(x) = sin(x) + x²

to:

f’(x) = sin(x + dx) + (x + dx)²

with numbers (plugging in 1.5 for “x”, it’ll look like this:

f(x) = sin(1.5) + (1.5)²

to:

f’(x) = sin(1.5 + dx) + (1.5 + dx)²

visually, it’ll look like this:

thx grant)

and we can give the “sin(1.5 + dx) + (1.5 + dx)²” part new labels, ’cause it’s quite long.

but we know that the equation above is the change in value due to the nudge we took…

so we call “sin(1.5 + dx)” → “d(sin(x))” & “(1.5 + dx)²” → “d(x²)”

“d(sin(x))*” & “d(x²)*” is the gap between where the original function ends & the nudged function ends. (look at the graph carefully)

* the “d” in front of the functions means to take the derivative, to find out the distance between those nudges.

“d(sin(x))” & “d(x²)”, when we take the derivative is…

f’(x) = cos(x) * dx & 2x * dx

f’(x) means the change of the original function due to the addition of “x” new value.

or u can call f’(x) as df, if that’s better for ya.

df = cos(x) * dx + 2x * dx

now, to get rid of the “dx”s, divide them out… and we get:

df / dx = cos(x) + 2x

the new function (aka the ‘adding’ function; df) divided by the small change in the “x” value (dx) gave us the sum of both changes* to the two graph lines (photo above).

* the words “changes” & “derivatives” are being used synonymously.

ok, time for the product rule.

f(x) = sin(x) * x² = Area of a rectangle

now, let’s add some extra width & height to this rectangle. just a bit more.

in real life, that’s how much we would have added.

but some of u can barely see that the yellow rectangles have a height & width.

and that’s crucial to understanding why we use the derivative to find out the area of the two thin, long rectangles.

so zooming in on the yellow blocks…

those thin pieces with the “d(x²)” & “d(sin(x)), that is the nudge we added onto the original figure.

and because we need to find the area, we will multiply the original width & height by the small nudges added to width & height.

it will look like this:

df = (x² * d/dx [sin(x)] ) + (d/dx [x²] * sin(x))

and when we take the derivatives…

df = (x² * cos(x) * dx) + (2x * dx * sin(x))

take the dx out of the right side of the equation into the left, we get:

df / dx = (x² * cos(x)) + (2x * sin(x))

the new function (aka the ‘adding’ function; df) divided by the small change in the “x” value (dx) gave us the sum of both changes* to the two graph lines (photo above).

* the words “changes” & “derivatives” are being used synonymously.

ok, now to combine functions inside of one another.

function composition, if we wanna be wise guys lol

this is the example:

g(x) = sin(x) ||| h(x) = x²

but now, what would happen if we wanted to put h(x) inside g(x)?

g(h(x)) = sin(x²)

here, we put “x²” wherever u see an “x” in the g(x) function.

and vice versa, it’ll look like this: h(g(x)) = sin²(x)

and here, we put “sin(x)” wherever u see an “x” in the h(x) function.

but we’ll focus on g(h(x)) for now.

and let’s say we nudge sin(x) to the left, and x² to the right, we will notice changes within how each side of the equation behaves.

now, to take the derivative & find out the value of the nudge for both x² & sin(x)… but combined together into one function.

g(h(x)) = sin(x²)

g(h(x)) will become d/dx, as we’re only looking for the change to the “x” value after the nudge in value.

d/dx = cos(x²) * 2x

yup, that’s it.

“sin” was the outer layer of the function. we must get the derivative for that.

then we must do the inside layer, “x²” separately, because sin(x) & x² are two separate functions, but squished into one.

this combined, composite function, is the new function (aka the ‘adding’ function; df) divided by the small change in the “x” value (dx) gave us the sum of both changes* to the two graph lines (photo above).

* the words “changes” & “derivatives” are being used synonymously.

CJ

© 2023–2100 by Carlos Manuel Jarquín Sánchez. All Rights Reserved.

--

--

No responses yet