The Chain Rule

Let $f(x)$ and $g(x)$ have derivatives, and consider the composite function $f(g(x))$. We shall develop a rule for use in differentiating this function, and this rule will greatly extend the collection of functions which we can differentiate. For example, the function $y=\sqrt{x^2+1}$ can be written as $f(g(x))$, where $f(u)=\sqrt{u},g(x)=x^2+1$, and we can differentiate each of these functions.

The difference quotient we must consider is

$$\frac{f(g(x+h))-f(g(x))}{h}.$$

If we consider the range of $g(x)$, that is, the set of values of $g(x)$, as a set of numbers of which we denote the general one by $u$, we first consider the difference

\frac{f(u+k) - f(u)}{k} - f'(u) .

For $k$ small, this expression is close to 0. We put it over a common denominator and write

\epsilon(k) = \frac{f(u+k) - f(u) - kf'(u)}{k} .

Then $ϵ(k)$ is a number depending on $k$, and is small when $k$ is small. Of course, $k=0$ is forbidden. Let us rewrite this equation as

f(u+k) - f(u) = k \epsilon(k) + k f'(u) .

If we were to choose any number at all and call it $ϵ(0)$, this last equation would be true for all $k$, including $k=0$. But in order to make $ϵ(k)$ small with $k$, we decide to choose the number 0 as the one which we shall call $ϵ(0)$. Thus we define $ϵ(0)=0$.

In the original difference quotient, let us write $u=g(x)$, $u+k=g(x+h)$. Then $k=g(x+h)-g(x)$. By the continuity of $g(x)$, $k\to 0$ as $h\to 0$, and so $ϵ(k)\to 0$ as $h\to 0$. Now we have

\frac{f(g(x+h)) - f(g(x))}{h} = \frac{f(u+k) - f(u)}{h} = \frac{k \epsilon(k) + k f'(u)}{h}= \frac{g(x+h) - g(x)}{h} \cdot \epsilon(k) + \frac{g(x+h) - g(x)}{h} \cdot f'(u) .

The quantities involved approach the limits indicated as $h\to 0$. Therefore the whole expression approaches f'(u)g'(x), and we have the chain rule:

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{(f(g(x)))' = f'(g(x)) \cdot g'(x).}

Exercise