Understanding Derivatives: My Struggle

[superdave]

I don't understand it. How it works, etc...

I've read the book definition, looked it up on the internet, and I still don't get it.

Like the current problem I'm working on:

derivative of sin((cos x)^2)*cos((sin x)^2)

 

[radou]

Read it again. Then try to identify 'what is what' in your problem.

 

[matt grime]

Here's the theory.

If f is a function, of x, what is the derivative of f (when/if it exists)?

It is another function df satisfying the property, that for small e

f(x+e)=f(x)+e*df(x) + o(e^2)

o(e^2) signifies the remainder, and it behaves quadratically, or worse, in e.

Now, it is easy to see what the derivative of f(g(x)) ought to be, if we are blase about things like e^2 and smaller terms. And there is no reason not to be so we can 'see' what is really going on.

If we add e to x, and if f and g are differentiable:
f(g(x+e) = f(g(x)+ e*dg(x) + o(e^2)) = f(g(x)) + e*dg(x)df(g(x)) + o(e^2).Now, look at your function. It is a product of two functions of functions. We can differentiate that just using all the rules we've learnt, so do so. Remember, the general case helps you understand what you're doing. The examples help you practice it.

 

[superdave]

Well I got this:

cos(sin^2 x) * 2cosx *-sinx*cos(sin^2 x) + -sin(sin^2 x) * 2cosx * cosx *sin(cos^2 x)

the back of the book gives me -sin2x cos(cos2x)

 

[matt grime]

You know your trig identities, right? (though I think you might be a little out in your differentiation.)

sin(2x)=2sin(x)cos(x), and so on.

 

[superdave]

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?

 

[JasonRox]

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.