Chain Rule

Stage: 5
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. To practise applying the Chain Rule, try our Mathmo App where you can generate plenty of questions and then check your answers.


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.

Notes and background
This explanation of the chain rule is fine for the sort of functions you will meet at A level. If you go on to study mathematics at university, or read some advanced calculus texts, you will see more rigorous definitions that do not rely on functions behaving suitably 'nicely'. Perhaps you might like to look through this explanation carefully, and see if you can uncover some of the hidden assumptions that we make!