superdave said:
of course I'm out on my differention. This textbook is written in the most obscure way possible. I missed the lecture where we went over this. My head really hurts. I'm malnurished because college dining halls care more about being cheap than healthy.
And I'm supposed to be a physics major. But if I can't understand this, how am I supposed to understand more complex math and physics?
So, you can live without understanding the Chain Rule today. Worry about the mathematics you are doing now, for now, and worry about the mathematics of tomorrow, for tomorrow. You aren't there yet, so why stress about it
It basically goes like this...
h(x) = f(g(x))
So, h(x) is like a function with a function inside of it. See it? We have g(x) "inside" f(x).
What's the derivative of h(x)? Well, the textbook should say...
h'(x) = g'(x)*f'(g(x))
So, the derivative of h(x) is simply the derivative of "inside" function multiplied with the "derivative" of f(x) then we put g(x) back "inside" of f(x).
Note: I use quotes because it isn't formal. My intention is only to show you how to use it, and maybe later you will understand it.
Here is an example:
h(x) = (x+x^2)^2
So, your g(x) is the "inside" function, which is g(x) = x+x^2. Your f(n) is the outside function, which is f(n) = n^2, where n=(x+x^2)=g(x). Now, you understand, why I used n for f(n) instead of x here.
So, find the derivative of h(x).
Using the formula...
h'(x) = g'(x)*f'(g(x))
g'(x) = 1+2x *If you don't know this, you have bigger problems.
f'(n) = 2n
So, input that in the formula and we get...
h'(x) = (1+2x)*2n = (1+2x)*2*(x+x^2)
Note: g(x) = n
And, we are done.