Why does it work?

This page is for those curious about how the chain rule works.  If you are interested, read slowly and carefully with pencil in hand.

Now that we know how to use the chain, rule, let's see why it works. First recall the definition of derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x},$$ where $\Delta f = f(x+h)-f(x)$ is the change in $f(x)$ (the rise) and $\Delta x=h$ is the change in $x$ (the run).

From change in $x$ to change in $y$

In other words, whenever $\Delta x$ is small, $\frac{\Delta f}{\Delta x}$ is close to $f'(x)$. Another way of saying that is: $$ \hbox{if } \Delta x \approx0, \,\, \text{ then }\, \Delta f \approx f'(x) \Delta x.$$ Multiplying by $f'(x)$ converts changes in $x$ into changes in $f(x)$. We say that $f'(x)$ is a conversion factor from changes in $x$ to changes in $f(x)$.

From change in $x$ to change in $y$, passing through $u$

Now suppose that $y=f(u)$ and that $u=g(x)$. Then $\frac{dy}{du}=f'(u)$ is the conversion factor between changes in $u$ and changes in $y$. That is, $$\Delta y \approx f'(u) \Delta u.$$ Meanwhile, $\frac{du}{dx}=g'(x)$ is the conversion factor between changes in $x$ and changes in $u$. That is, $$\Delta u \approx g'(x) \Delta x.$$

Put those two results together to get $$\Delta y \approx f'(u) \Delta u \approx f'(u)g'(x) \Delta x,$$ which means that the conversion factor from $\Delta x$ to $\Delta y$ is $f'(u)g'(x)$, or equivalently that $$\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}.$$ That's the chain rule!