Calculus is the study of changing variables.

Important Symbols

$d i f$ The differential symbol. Means a “little bit of”. So, $d i f x$ means a “little bit of $x$ “. Could be infinitely small.
$in t e g r a l$ The integral symbol. Long s means “sum of”. So, $in t e g r a l d i f x$ means the “sum of the little bits of $x$ “

On smallness

At a certain point, small values become negligible. Consider a square where $x$ represents the length of one side. The area of the square, therefore, is $x^{2}$ . Now, consider that you were to add $d i f x$ to $x$ , meaning increasing each side of the square by $d i f x$ . How much would the area grow by? This can be represented by the following diagram: By taking the area with $x + d i f x$ , we get $(x + d i f x)^{2}$ , which can be expanded to: $x^{2} + 2 d i f x + d i f x^{2}$ . Now, note that the $d i f x^{2}$ term is an order of magnitude smaller than even $2 d i f x$ . Because it is so small, we are free to ignore it and consider it to be effectively zero. This is even more true for smaller and smaller values of $d i f x$ .

On relative growings

Calculus is largely about how different values can grow and change in relation to one another. Consider the following example: Let $x$ represent the horizontal distance, from a wall, of the bottom end of a ladder $" A B "$ . Let $y$ be the distance at which the top of the ladder rests on the wall. The question arises: If we were to pull the ladder further away from the wall, thereby increasing $x$ , how much would $y$ change? The following figure represents the situation: We will solve this with example data first. Suppose the ladder was so long such that where $A$ is $19" in c h es "$ from the wall, $B$ meets the wall $15" f t "$ above. If $x$ changes by $1" in c h "$ , then, how much does $y$ change by?

#let x = 19
#let y = 180
#let dx = 1
#let L = calc.sqrt(calc.pow(x, 2) + calc.pow(y, 2))
#let z = (calc.pow(L, 2) - calc.pow(x + dx, 2))
#let dy = (y - (calc.round(calc.sqrt(z), digits: 3)))
$ x = #x "in" \
y = #y "in" \
dif x = #dx "in" \
dif y = "?" \ $
Because $y$ must decrease, we know that $h$, the new height, is:
$ h = y - dif y $
And, conversely, the distance from the wall, $l$, is:
$ l = x + dif x $
By pythagorean theorem, we know the length of the ladder is:
$ sqrt((#y)^2 + (#x)^2) = #L $
Squaring both sides gets us:
$ #y^2 + #x^2 = #L^2 $
Becuase the length of the ladder must remain constant, the new \
height must be:
$
(y - dif y)^2
= (#L)^2 - (#(x + dx))^2
= #(calc.pow(L, 2)) - #calc.pow(x + dx, 2)
= #z
$
Taking the square root,
$ y - dif y = #(calc.round(calc.sqrt(z), digits: 3)) $
So, $dif y$ is:
$ #(calc.round(dy, digits: 3)) $
 
This makes the ratio at which y changes in relation to x, or:
$ (dif y) / (dif x) = -#(calc.round(dy, digits: 3)) / #dx $

The above example also delivers a key intuition: in order for $(d i f y) / (d i f x)$ to make any sense, y and x must have some relation. In this example, the relation was the following:

L = s q r t (y^{2} + x^{2}) y = s q r t (L^{2} - x^{2})

So, because there is a relation, a change in x affecting y by some amount makes sense.

Simplest Cases

Power Rule

Power rule is defined as the following:

y = x^{n} (d i f y) / (d i f x) = n x^{(} n - 1)

I will not provide a formal proof for this rule now, just an example as to how it may work internally:

Consider
$ y = x^2 $
We know that if $x$ grows, $x^2$ grows, so $y$ grows
$ y + dif y = (x + dif x)^2 $
By expansion,
$ y + dif y = x^2 + (2x dif x) + (dif x)^2 $
By the discussion on smallness, $(dif x)^2$ is simply too small to matter \
Moreover, we said before $y = x^2$. Let us substitute:
$ x^2 + dif y = x^2 + (2x dif x) $
Simplifying,
$ dif y = 2x dif x $
So,
$ (dif y) / (dif x) = 2x $

Sources

Much of this note comes from one source: Calculus Made Easy by Silvanus P. Thompson This is an attempt to intuitively explain fundamental concepts of calculus rather than going through rigorous math.

notes

Basics of Calculus