Calculus for the Curious

Preface

“What we know is a drop, what we don't know is an ocean.”
- Isaac Newton

Welcome! What you're reading right now is an introduction to the basics of calculus. I wouldn't be surprised if you already have a preconceived negative opinion of the subject.
But fear not!
Calculus is actually rather easy, as long as you have a very good understanding of the very basics.
And the aim of this site is to provide you with exactly that!
I encourage you to take a look at the first chapter, even if this seems outside your current interests or abilities.
This text should only require a basic (but thorough!) level of knowledge of algebra, and an understanding of simple functions.

1.What speed are we going at?
2.The definition of the derivative
3. Derivatives of simple functions
4.Properties of the derivative
And beyond!

What speed are we going at?

Imagine this (rather simple) scenario:
- you're in a car.
- you look at the speedometer.

What does the speedometer display? This question may seem trivial, but bear with me.
Your speed, or more accurately, your velocity.
You probably already know the formula for (and definition of) velocity: $$v = \frac{s}{t}$$

You might be already seeing the problem here.
You're looking at the speedometer at a single point in time.
Does that mean that $t=0$?
But we can't divide by 0!
Now, hold that thought for a minute.

The Delta.

The speed equation above, while not inacurate, omits a pretty big point.
Usually, the simpler definition suffices, but we want to make it clear what's happening.
What you're measuring when using it, is the difference in speed and velocity.
For example:
If you want to average velocity of a car (assuming you know the length of the track) with a clock, you wouldn't use the time the car finished the race,
but instead you would subtract the hour it finished from the hour it started.

Mathematicians use a delta - $\Delta$ (from "difference") to signify that.
so, whenever you see a $\Delta$ in a formula, interpret it as: $$\Delta x = x_2 - x_1$$ where $x_2$ is x at a prior moment, and $x_2$ is x at a later moment.
In simple terms, it is "the change in x"

So now, to restate our velocity formula: $$v = \frac{\Delta s}{\Delta t}$$ But that doesn't help us at all! We're looking at the speedometer in a singular moment of time, so: $$t_1 = t_2$$ $$\Delta t = 0$$ And we run into the exact same problem again.

Let's try to approximate it!

Our equation gives us the average velocity over a period of time.
If you were to drive 100 kilometers in an hour, your average velocity would be, well, 100 kilometers per hour.
But chances are your speedometer wasn't displaying 100 km/h all the time.
You slowed down on the turns, sped up on an empty highway, stopped to let someone cross the road.

Your speed varied a lot during the ride, so this clearly wouldn't be a great approximation for every single moment.
But, over smaller periods of time, there weren't any drastic changes.
Sure, you might've braked really hard after seeing a cat on the road, but if we make that into an even smaller, shorter fragment, the change will also be accurate.

Ok, let's try it!
You notice a counter, showing you how many kilometers the car has traveled. Convenient!
So, you record it using your phone, along with the speedometer.
You use the aforementioned formula to compute your velocity.
At first, when the period is longer, your calculations don't exactly match up with the speedometer. But, as you use shorter and shorter periods of time, they start to show the same.

What is going on here?

Truth is, what your speedometer is doing, is exactly that. Computing the velocity over a really short time.
Depending on your car model, it might be so fast you wouldn't even notice it.

This prompts more questions than it resolves.
If your computations got closer and closer to the speedometer, what if the speedometer would use an even shorter period of time? Is there any definite way to compute this? There is always a number closer to zero.
0.1, 0.01, 0.001, and so on.

The definition of the derivative

The derivative is closely related to what we were thinking about just now.
Here is its definition, though i warn you: it is missing a very important part. $$f'(x) \approx \frac{ f(x + h) - f(x)}{h}$$ for small values of h.
$f'(x)$, as you might've guessed, is a common notation for the derivative of x, that we will be using here.
What this formula is saying, is in essence: "if we increase x by h, how much does the value of the function increase by in proportion to h?

Let's treat the road of the car $s$ as a function of time $s(t)$, since it changes depending on it.

Exercise:

Try and merge these two formulas, to get the "derivative" of s.

$$s'(t) \approx \frac{ s(t + h) - s(t)}{h}$$ for small values of h.

That is essentially the formula we were using earlier! But these are only approximations.
What if we want the exact formula?
We could see the approximation gets better as h gets closer to zero, but how do we express that?

Limits to the rescue!

We won't be going in-depth into limits here, that isn't generally required for understanding the derivative.
What you need to know is just the fact that the limit of a function as h goes to a number n, written as $$\lim_{h \to n}$$ is the function, but with h "infinitely close" to the number. It is strictly not the same as the function without the limit.
You should mostly operate as if h is just a really slightly, negligibly different from the number. If you wish to learn more about what that actually means (rigorously), search for the "Epsilon Delta definition of a limit".

Exercise 2:

What does this expression equal? $$\lim_{x \to 0} \frac{1}{x}$$

We can see that, for smaller and smaller values of x, the function grows. $$\frac{1}{0.1} = 10$$ $$\frac{1}{0.01} = 100$$ $$\frac{1}{0.001} = 1000$$ So, we conclude: $$\lim_{x \to 0} \frac{1}{x} = \infty$$ You might not be exactly used to an infinity appearing in your equation, but don't worry, it's quite the common occurrence.

So with this tool, we can finally state the exact definition of the derivative in full: $$f'(x) = \lim_{h \to 0} \frac{ f(x + h) - f(x)}{h}$$ This might not seem useful in any way, but when we determine the derivative of mathematical functions, it will allow us to get exact values of the derivatives. By now, we've gained some insight about what derivatives describe.
Much like velocity is the rate of change of the distance in time, the derivative is the rate of change of the function, as x changes.

Derivatives of simple functions

Let's say our car's position always is a constant: $s(t) = c$, for all values of t. What would be the derivative (and thus velocity) then? You should see that $\Delta s = 0$, But try to work this out using the formula.

Exercise 3

Compute the derivative of the function.

$$s'(t) = \lim_{h \to 0} \frac{ s(t + h) - s(t)}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ c - c}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{0}{h}$$ since the h no longer does anything (zero divided by any number is always zero), we can remove it from the limit! $$s'(t) = 0$$

Now our car is moving proportionally to the time (at a constant pace). $$s(t) = at$$ Once again:

Exercise 4

Compute the derivative of the function.

$$s'(t) = \lim_{h \to 0} \frac{ s(t + h) - s(t)}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ a(t + h) - at}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ at + ah - at}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ah}{h}$$ and since $\frac{h}{h}$ will always equal one: $$s'(t) = \lim_{h \to 0} a$$ $$s'(t) = a$$ so, if $s(t) = at$, $s'(t) = a$

Makes sense! If our car is moving an unit distance in unit time, that should be our velocity.

Now, try and differentiate (compute the derivative of): $$ s(t) = t^2$$

Exercise 5:

$$s'(t) = \lim_{h \to 0} \frac{ s(t + h) - s(t)}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ (t + h)^2 - t^2}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ t^2 + 2th + h^2 - t^2}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ 2th + h^2}{h}$$ $$s'(t) = \lim_{h \to 0} 2t + h$$ $$s'(t) = 2t$$

And, Last but not least,
What if the function looks like this: $$s(t) = \frac{1}{2}a{t^2}$$ You may recognize this as the equation for uniformly accelerating motion!
So our result here will be the velocity of this motion at any point in time.

Exercise 6:

$$s'(t) = \lim_{h \to 0} \frac{ s(t + h) - s(t)}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ \frac{1}{2}a(t + h)^2 - \frac{1}{2}at^2}{h}$$ Using the square of a sum formula: $$s'(t) = \lim_{h \to 0} \frac{ \frac{1}{2}a(t^2 + 2th + h^2) - \frac{1}{2}at^2}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ \frac{1}{2}at^2 + \frac{1}{2}2ath + \frac{1}{2}ah^2 - \frac{1}{2}at^2}{h}$$ $$s'(t) = \lim_{h \to 0} \frac{ ath + \frac{1}{2}ah^2}{h}$$ $$s'(t) = \lim_{h \to 0} at + \frac{1}{2}ah$$ $$s'(t) = at$$

The Power Rule

There does exist a rule for functions of form $x^n$.
Proving that is outside the scope of this text, but just know that: $$(x^n)' = nx^{n-1}$$ For the proof, go to here

Properties of the derivative

Computing derivatives manually, using the definition like this, is definitely a tiring task. Thankfully, we can identify some properties, that make this much easier.

Adding Derivatives

The general formula for adding two derivatives is quite simple! $$(f(x) + g(x))' = f'(x) = g'(x)$$ This fact is probably quite obvious, but feel free to try and prove it if it seems counterintuitive. The same follows for subtraction:

Derivative of a function multiplied by a constant

Another quite simple formula: $$(cf(x))' = cf'(x)$$ This directly follows from the definition: $$(cf(x))' = \lim_{h \to 0} \frac{cf(x + h) - cf(x)}{h}$$ $$(cf(x))' = \lim_{h \to 0} \frac{c(f(x + h) - f(x))}{h}$$ And, since the constant is not affected by the limit: $$ \begin{align} (cf(x))' &= c(\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}) \\ & = cf'(x) \end{align} $$

And beyond!

Thank you for reading!
By now you hopefully have acquired some understanding of how derivatives work, why they exist, and what laws they're governed by What now?
Well, we learn further, of course! Here are some resources I personally enjoy:

- The art of problem solving Series
- 3Blue1Brown (Youtube)
- Math is Fun
- Math - Libretexts