You've heard it before: "You need to know calculus to understand this", or "You can't do that unless you know calculus", or "You must worship calculon, the Greek god of calculus", or "You can't urinate in here sir, this is a tool shed", etcetera. Good news! Today, you'll be a step closer to telling those people to stuff it!

Today, I'm going to cover limits, deriving using delta method, simple rules to derive a variety of equations. I'll try to provide an example for everything.

1. Limits

In Limits, you're always trying to do math without actually doing math. If you did the question in the limit, what number would it be close to? That's the question a limit tries to solve.

lim    
x->2   x + 2 = 4

Intuitively, you know the answer will be 4, and as x approaches 2, the answer will approach 4, since 2 + 2 = 4. Meanwhile,

lim       1
x -> 0    - = Infinity
          x

is actually infinity, because right as you're about to hit zero, you'll be just one step away from infinity. In most classical calculus classes, you'll learn how to do limits for all sorts of impossible equations where you're dividing by zero, but in practice, that's not very useful, since even using the basic element of calculus we'll be covering in the next section.

2. Delta Method derivation

Before we start with the technique for delta method derivation, you have to understand what calculus is.

Calculus is just the slope of an equation. Anyone who has taken a High School Mathematics should be able to remember:

 m = rise/run = delta-Y/delta-X = Y2 - Y1 / X2 - X1

Derivative calculus is just trying to do the same. Also, like Slope m is the rate of change of Y each X, (every time X changes by 1, Y changes by m), the derivation of an equation is the rate of change of the Y in an equation for every X. Of course, calculus would be a pain if you could only use y and x to describe variables, so the notation for a derivitave is often:

dy
--
dx

Where y and x can be any letter. For example, a standard physics equation for linear velocity is:

x = x0 + v0t + 1/2at^2

where x0 is the initial position, where v0 is the initial velocity, a is acceleration and t is time. Expressing the derivitave (which becomes, incidentally, the equation for expressing velocity)

dx
-- = v0 + at
dt

This should illustrate the simplicity of the system to you, especially as we move on into more complex subjects where you'll begin to understand how the above works. I will be using the above equations again in later articles, because kinematics (equations like x = vt and v = at) provide a direct example of applied calculus.

Now, understanding the notation a little better, and the concept just a little, now we're going to look at our first derivitave, deriving the equation y = x^2 + 2x + 1. To do this, we'll use a technique known as the delta method.It's very simple once you get the hang of it, but it's the clumsiest way to do it, and mostly useful for demonstrating how a derivitave is done(lucky you!!).

It's the long four step method:

i. Add a delta to the X and Y in your equation. This is like adding a tiny tiny amount to it.

y + delta-y = (x + delta-X)^2 + 2(x + delta-X) + 1

ii. Take away Y from each side. This means taking y away from the left side of the equation and the eqaution for y from the right side.

y + delta-y - y = (x + delta-x)^2 + 2(x + delta-x) + 1 - (x^2 + 2x + 1)

We can multiply out 2(x + delta-x) to become 2x + 2delta-x, which lets us take out 2x and +1 , so we have:

delta-y = (x + delta-x)^2 + 2delta-x - x^2

Now we'll want to multiply out the exponent there, which expands to (x + delta-x)(x + delta-x), so our equation becomes

delta-y = x^2 + 2x(delta-x) + delta-X^2 + 2delta-x - x^2

so

delta-y = 2x(delta-x) + delta-x^2 + 2delta-x

iii. divide the answer by delta-x to get a dy/dx

delta-y
------- = 2x + delta-x + 2
delta-x

iv. finally, use the limits we did before to send delta-x towards zero. In practice, this generally just means setting all remaining delta-xes to zero.

delta-y
-------  = 2x + 2
delta-x
  lim delta-x->0

So the derivitave for x^2 + 2x + 1 is

dy
-- = 2x + 2
dx

And you've witnessed your first derivation!

3. The tricks to make calculus easy
i. Exponents
If you had to run out and do the above every time you wanted to do simple derivation, calculus would not be very popular among mathematicians, physicists, and engineers. Luckily, the delta method produces very simple, reproducable results when you learn the rules. For example: The equation

y = x^2

can be expanded by delta method to be the following. Note that I'm using dx and dy instead of delta-x and delta-y. This does not have any relation to the dy/dx notation, I'm just lazy and there's no delta key. Only the final answer uses dy/dx notation.


i.
y + dy = (x + dx)^2
y + dy = (x + dx)(x + dx)
y + dy = x^2 + 2x(dx) + dx^2

ii.
y + dy - dy = x^2 + 2x(dx) + dx^2 - x^2

iii.
dy
-- = 2x + dx
dx

iv.
dy
-- = 2x
dx
 lim dx->0

so

dy
-- = 2x
dx

This demonstrates the first rule of calculus:

dy
--(x^n) = nx^(n-1)
dx

That looks complicated, but look at the exampe we did the the delta method: x^2 became 2x^1. If we had an equation:

y = x

Then we can quickly derive it to be:

dy
-- (x^1) = 1
dx

or 1 * x^0 = 1.

Another important thing to remember at this point is that equations seperated by spaces don't effect each other. For example, if you have the equation:

y = x^2 + x^6 + x^1

You'd derive each stage of the equation independantly, so the equation easily derives to

dy
-- = 2x + 6x^5 + 1
dx

which, as you can see, doesn't require any thing other than three applications of the first rule of calculus.

a note about this rule, it's always the same, no matter if the exponent is positive, negative, or a fraction. This means that The equations

y = x
y = 1/x = x^-1
y = sqrrt(x) = x^(1/2)

derive to

dy
-- = 1
dx

dy
-- = -1x^-2
dx

dy
-- = (1/2)x^(-1/2)
dx

respectively.


ii. Constants
Constants by themselves in an equation are dropped when deriving. A constant multiplying a variable, however, is left in. For example, take the equation:

y = 2x^2 + 1

As you can see, there is a constant multiplying X, and a constant 1 sitting by itself. To derive these, we'll use the exponents rule for the first, and we'll just drop the second.

dy
-- = 4x
dx

It's just that easy!

iii. Functions multiplied together

Let's say we have the equation:

y = (x^2)(2x^3)

You can't derive this using the exponent rules as seen above. To derive this equation, you must take the derivitave of one multiplied by the non-derivitave of the other, then add the the derivitave of the other multiplied by the non derivitave of the first. Our equation becomes:

dy
-- = (x^2)(6x^2) + (2x^3)(2x)
dx

If you were to simplify (x^2) to u and (2x^3) to v, it becomes much clearer as:

dy
-- u(dy/dx v) + v(dy/dx u)
dx

iv. Functions divided by another function

Now let's say we have an equation:

y = (x^2)/(2x^3)

Again, we can't use any of the rules we've seen up until now, so we use another method again. Simplifying again (x^2) to u and (2x^3) to v, the way you derive functions divided into one another is:

dy   v(dy/dx u) - u(dy/dx v)
-- = -----------------------
dx             v^2

and applying that to our example:

(2x^3)(2x) - (x^2)(6x^2)
------------------------
     (2x^3)^2

which can be simplified as appropriate.

Congratulations, you have made it through the fundamentals of calculus! The mighty calculon has been defeated, and you can take a leak in that tool shed! Play with these techniques with your favourite equations and see what you can come up with!

--SJ Zero was dressed like dorothy when he found the guy behind the curtain.