What Is an Understandable Proof of the Chain Rule?

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Homework Statement



I'm looking for a proof for the chain rule that is relatively easy to understand. Can someone show / link me one? Thanks.

Homework Equations





The Attempt at a Solution

 
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Google is your friend:

http://math.rice.edu/~cjd/chainrule.pdf
 
I posted my calculus notes with a basic (non-rigouous) derivation of the chain rule in, you could have a look at those.
 
For the pfd file, I don't understand the middle of page 2 where it says

"..and use the second equation applied to the right-hand-side with k =
[g'(x) + v]h.. Note that using this quantity for k tells us
that k -> 0 as h -> 0, and so w -> 0 as h -> 0."

How did they choose that substitution for k?
 
anyone?
 
how about this one?
let u(x) be a differentiable fuction in [a,b] with values in [a',b'] and y=f(x) a differentiable fuction in [a',b'].
Let Δx be a randomly picked difference x2-x1. That causes a change Δu on u(x), while Δu causes a change on y=f(u).
We have Δu=(u'(x)+n1)*Δx. You can easily verify by looking at the graph that the line connection the points (x1,u(x1)) and (x2,u(x2)) has a slope equal to the value of the derivative of u on x1 plus a number n1 to compensate for the fact that Δx isn't zero(and thus this line isn't the tanget on x1).
The same applys to Δy=((f'(u)+n2)*Δu.
When Δx-->0 , n1,n2-->0
Δx/Δy=(f;(u)+n2)*(u'(x)+n1)
We calculate the limit of the fraction when Δx-->0 and it is equal to f'(u)*u'(x)=(dy/du)*(du/dx)
system has gone crazy and won't show the math symbols, sorry for the formating

It's the proof from Louis Brand's book "Advanced Calculus", paragraph 52- The chain rule
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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