Chain Rule

This short article shows one way of deriving the Chain Rule for differentiation. Read it carefully and see if you can follow the argument. You may wish to try to recreate the proof for yourself. 

Suppose we have differentiable functions $f(x)$ and $g(x)$. 

The derivative of $f(x)$, written $f'(x)$, is given by $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\;.$$

Suppose we want to differentiate the composition of functions $f(g(x))$.

Then by our definition of the derivative, $$\frac{d}{dx} \left( f(g(x)) \right) = \lim_{h\to 0} \frac{f(g(x+h))-f(g(x))}{h}\;.$$

In order to make sense of this, we're going to need to introduce a couple of new variables:

$$

\begin{align}

    k &= g(x+h) - g(x)\quad\mbox{and}\\

    u &= g(x)\;.

\end{align}

$$

Now we can rewrite the above:

$$\eqalign{\frac{d}{dx} \left( f(g(x)) \right) &= \lim_{h\to 0} \frac{f(u+k)-f(u)}{h} \cr &= \lim_{h \to 0} \frac{f(u+k)-f(u)}{h}.\frac{k}{k} \cr &= \lim_{h \to 0} \frac{f(u+k)-f(u)}{k}.\frac{g(x+h)-g(x)}{h}}$$

We know that as $g$ is differentiable, it is also continuous, so $k=g(x+h)-g(x) \to 0$ as $h \to 0$.

So, as the limit of the products is the product of the limits, the right hand side of our equation becomes $f'(g(x))g'(x)$ in the limit.