how to prove the chain rule? I have just learnt about the chain rule but my book doesn't mention a proof on it. I tried to write a proof myself but can't write it. So can someone please tell me about the proof for the chain rule in elementary terms because I have just started learning calculus.
Assuming everything behaves nicely ($f$ and $g$ can be differentiated, and $g(x)$ is different from $g(a)$ when $x$ and $a$ are close), the derivative of $f(g(x))$ at the point $x = a$ is given by 
$$
\lim_{x \to a}\frac{f(g(x)) - f(g(a))}{x-a}\\ = \lim_{x\to a}\frac{f(g(x)) - f(g(a))}{g(x) - g(a)}\cdot \frac{g(x) - g(a)}{x-a}
$$
where the second line becomes $f'(g(a))\cdot g'(a)$, by definition of derivative.
If I understand the notation correctly, this should be very simple to prove:

Assume $f(x) = g(h(x))$

Then:
$$f'(x) = \frac{df(x)}{dx}$$

This can be expanded to:
$$\frac{df(x)}{dx} = \frac{df(x)}{dg(h(x))} \frac{dg(h(x))}{dh(x)} \frac{dh(x)}{dx}$$

When you cancel out the $dg(h(x))$ and $dh(x)$ terms, you can see that the terms are equal.

Since $f(x) = g(h(x))$, the first fraction equals 1.
($$\frac{df(x)}{dg(h(x))} = 1$$)

If we substitute $h(x)$ with $y$, then the second fraction simplifies as follows:
$$\frac{dg(y)}{dy} = g'(y)$$
Substituting $y = h(x)$ back in, we get following equation:
$$\frac{dg(h(x))}{dh(x)} = g'(h(x))$$
The third fraction simplifies to the derrivative of $h(x)$ with respect to $x$. This can be written as
$$\frac{dh(x)}{dx} = h'(x)$$

Substituting these three simplifications back in to the original function, we receive the equation

$$\frac{df(x)}{dx} = 1g'(h(x))h'(x) = g'(h(x))h'(x)$$

This is the chain rule.