[high School - Trigonometry] Don't know how derivatives work

[Professor Raine]

Okay, due to my Senioritis, I didn't pay attention to the beginning of the class, which taught us about derivatives and how they work. Please explain how they work, give a definition, and an example on how they work.

[Squeek]

I don't know what they are, but I know what they do.

Basically, the formula for solving derivatives is:

derivative of x^n = n*x^n-1

So x^2 = 2x. x^3 = 3x^2. 5x^2 = 10x.

If you're asked to bring a derivative down until there's no exponent, you do this:

x^4 = 4x^3 = 12x^2 = 24x.

Oh, and the derivative of X is 1. There are a lot of basic derivatives that you have to memorize. Which is why I failed Calculus.

[flipsta_lombax]

Well, here is Wiki: http://en.wikipedia.org/wiki/Derivative

I never learned derivatives until now in Calculus; I never knew you could learn it in Trig. But yeah, dude, this took almost a semester to cover, so explaining would be just too long.

All I can tell you at the moment is that the derivative is a rate of change in a function. Everything else is in that link. ;P

[Benny1]

Calc in Trig? Unexpected, but okay.

The derivative of a function, is as has been said, the immediate rate of change of a function.

The basic way to derive it is this. Slope of a function is f(x1) - f(x2) / x1 - x2. Now, let's say we're doing... f(x1 + h) - f(x1) / (x1 + h) - x1. This is quite reasonable, and all. Now, I really hope you know limits. take that, as lim h -> 0. Basically, work everything out, and it SHOULD work out that you have a legit number. If you don't, the function isn't differentiable at that point, but whatever.

Tricks to this involve this: The basic idea is the derivative of X^n = nX^(n-1). Addition rule, basically, if you have 5x + 5x^2 or something, just take the derivative of each one and add it, so 5 + 10x. Subtraction is the same. For multiplication, it's more complex. For say, 5x^2 * 2y^3, take derivative of first times the second, add the derivative of the second times the first to that. For division, it's, for low/high, lowd(high) - highd(low) / low^2. d(function) means derivative.

There's chain rule and stuff, but that'd take more time. Want that explained?

[Relambrien]

Okay, I'm assuming you know about the slopes of lines and all that. Basically how fast the function's value changes. For instance, in the equation f(x) = 4x + 5, the slope is 4. For every increase in x of 1, y increases by 4. But you can only take the slope of straight lines.

Derivatives remove this limitation, allowing you to take the "slope" of nearly any function at nearly any point. They allow you to find out how fast the function's value is changing at any one instant.

But you may ask, "If you're only looking at an instant, how can you determine how fast the function is changing? Because at any instant, the function won't change at all." This is true. The slope of a line is given by the change in y over the change in x, which at any instant gives you the indeterminate form 0/0.

We solve this problem by using limits. Limits are a value a function approaches but never truly reaches. For instance, if you have the function f(x) = 1/x, you'll see how as x gets greater and greater, y gets closer and closer to zero but never truly reaches zero. Thus, the limit of 1/x as x approaches infinity is -equal- to zero.

The derivative is defined as the limit, as h approaches zero, of (f(x + h) - f(x))/h. In this case, h is some arbitrary value that you add to x, so that you can get two points. If you break that equation down, what you're really left with is this:

f(x + h) - f(x) is essentially the change in y. You take f(x) at some point, then subtract it from the value of f(x + h), which is your second point.

Then you divide all of that by h, which is really (x + h) - x, or the change in x. That whole equation is really just the change in y over the change in x, but it makes the changes get progressively smaller. You can pick any value of h that isn't zero and you'll get a real answer, so all you have to do is evaluate that expression at progressively smaller values of h, until you find the value the answers seem to be approaching, but never reach.

For instance, if you got 5.5, 5.25, 5.125, 5.0625, etc., then you could reasonably assume that the derivative at that point is 5. The numbers appear to be approaching 5 but never reach it.

However, a better way to use that expression is to find a general solution for the derivative at -any- point of a function. For instance, say you want to find the derivative of f(x) = x^2.

Note, the expression "dy/dx" simply means "the derivative of y," and is pronounced in varying ways. You can say "dee why dee eks," "the derivative of y with respect to x," or any number of other things.

dy/dx = lim h>0 (f(x + h) - f(x))/h

Now we sub in the function we want

dy/dx =lim h>0 ((x+h)^2 - x^2)/h

When you expand this, you get

dy/dx = lim h>0 (x^2 + 2hx + h^2 - x^2)/h
dy/dx = lim h>0 (2hx + h^2)/h

It's easier to see what happens next if you separate the remaining expression like this

dy/dx = lim h>0 (2hx/h + h^2/h)

Now here's where things get fun. Because we're taking the -limit-, and h isn't exactly equal to zero, we can cancel it out. If this wasn't the limit, and h was equal to zero, we'd be dividing by zero and that isn't possible. The limit is what allows us to do this.

dy/dx = lim h>0 2x + h

In 2hx/h, you can cancel out the two h terms. In h^2/h, you can cancel one out, leaving you with h/1, or just h.

Now here's the -best- part. Because of the limit, although h -isn't- zero, for all intents and purposes it may as well -be- zero. Therefore, we can take it out now as if it were.

dy/dx = 2x

That's the derivative of x^2.

Let f(x) = x^2

df(x)/dx = 2x

The df(x)/dx just means the derivative of f(x) with respect to x, and we defined f(x) to equal x^2.

What this allows you to do is take the instantaneous rate of change of the function x^2 at any point. For instance, at x = 4, the derivative becomes 2(4), or 8.

Now, the good thing about derivatives is that you don't have to use that big nasty (f(x +h) - f(x))/h expression, known as the difference quotient, very often. As you go through and learn more about derivatives, you'll develop a sort of library of rules you can use to quickly calculate the derivative of nearly anything.

For instance, you'll learn that if you have a function f(x) = x^n, where n is any real number, that df(x)/dx = nx^(n-1). So the derivative of x^3 is 3x^2. The derivative of x^(1/2) is (1/2)x^(-1/2). That last one is especially useful because it helps you calculate derivatives of square roots.

There are other rules you'll learn, one of the more powerful being the chain rule, but that'll come later.

FYI, the chain rule basically says that if you want to take the derivative of a composite function, you take the derivative of the outer function and multiply it by the derivative of the inner function. The derivative of f(g(x)) = (df(x)/dg(x))(dg(x)/dx). So if you have something like (4x + 1)^2, the derivative is (2(4x + 1))(4), or 32x + 8. But like I said, that's later.

So, to recap.

The derivative of a function is the instantaneous rate of change of the function, or how fast the function is changing at any single given point. They are calculated through the use of limits, which bypass the problem of being left with a 0/0 term when trying to take the slope using only one point. A commonly-used derivative in the real world is velocity. Velocity is the first derivative of position with respect to time. In other words, to find out how fast you're moving at any point, you take the derivative of the function representing your distance.

To calculate the derivative the hard way, you use the difference quotient outlined above. However, you'll learn a very large number of rules and principles that ensure you rarely ever need the difference quotient, and can instead calculate derivatives much more simply. These will come with time, as you progress into learning differential calculus.