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
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.
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,
x -> 0 - = Infinity
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:
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)
-- = v0 + at
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
delta-y = 2x(delta-x) + delta-x^2 + 2delta-x
iii. divide the answer by delta-x to get a dy/dx
------- = 2x + delta-x + 2
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
------- = 2x + 2
So the derivitave for x^2 + 2x + 1 is
-- = 2x + 2
And you've witnessed your first derivation!
3. The tricks to make calculus
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.
y + dy = (x + dx)^2
y + dy = (x + dx)(x + dx)
y + dy = x^2 + 2x(dx) + dx^2
y + dy - dy = x^2 + 2x(dx) + dx^2 - x^2
-- = 2x + dx
-- = 2x
-- = 2x
This demonstrates the first rule of calculus:
--(x^n) = nx^(n-1)
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:
-- (x^1) = 1
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
y = x^2 + x^6 + x^1
You'd derive each stage of the equation independantly, so the equation
easily derives to
-- = 2x + 6x^5 + 1
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)
-- = 1
-- = -1x^-2
-- = (1/2)x^(-1/2)
Constants by themselves in an equation are dropped when deriving. A
constant multiplying a variable, however, is left in. For example, take
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.
-- = 4x
It's just that easy!
iii. Functions multiplied
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:
-- = (x^2)(6x^2) + (2x^3)(2x)
If you were to simplify (x^2) to u and (2x^3) to v, it becomes much
-- u(dy/dx v) + v(dy/dx u)
iv. Functions divided by another
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)
-- = -----------------------
and applying that to our example:
(2x^3)(2x) - (x^2)(6x^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.