### Subsection2.9.2(Optional) — Derivation of the Chain Rule

First, let's review what our goal is. We have been given a function $g(x)\text{,}$ that is differentiable at some point $x=a\text{,}$ and another function $f(u)\text{,}$ that is differentiable at the point $u=b = g(a)\text{.}$ We have defined the composite function $F(x) = f\big(g(x)\big)$ and we wish to show that

\begin{align*} F'(a) &= f'\big(g(a)\big) \cdot g'(a) \end{align*}

Before we can compute $F'(a)\text{,}$ we need to set up some ground work, and in particular the definitions of our given derivatives:

\begin{align*} f'(b) &= \lim_{H \to 0} \frac{f(b+H)-f(b)}{H} & \text{and }&& g'(a) &= \lim_{h \to 0} \frac{g(a+h)-g(a)}{h}. \end{align*}

We are going to use similar manipulation tricks as we did back in the proofs of the arithmetic of derivatives in Section 2.5. Unfortunately, we have already used up the symbols “$F$” and “$H$”, so we are going to make use the Greek letters $\gamma, \varphi\text{.}$

As was the case in our derivation of the product rule it is convenient to introduce a couple of new functions. Set

\begin{align*} \varphi(H) &= \frac{f(b+H)-f(b)}{H} \end{align*}

Then we have

\begin{align*} \lim_{H \to 0} \varphi(H) &= f'(b) = f'\big(g(a)\big) & \text{since } b=g(a), \end{align*}

and we can also write (with a little juggling)

\begin{align*} f(b+H) &= f(b) + H \varphi(H) \end{align*}

Similarly set

\begin{align*} \gamma(h) &= \frac{g(a+h)-g(a)}{h} \end{align*}

which gives us

\begin{align*} \lim_{h \to 0} \gamma(h) &= g'(a) & \text{ and } && g(a+h) &= g(a) + h \gamma(h). \end{align*}

Now we can start computing

\begin{align*} F'(a) &= \lim_{h \to 0} \frac{F(a+h)-F(a)}{h}\\ &= \lim_{h \to 0} \frac{f\big(g(a+h)\big)-f\big(g(a)\big)}{h} \end{align*}

We know that $g(a) = b$ and $g(a+h) = g(a) + h \gamma(h))\text{,}$ so

\begin{align*} F'(a) &= \lim_{h \to 0} \frac{f\big(g(a) + h\gamma(h) \big)-f\big(g(a)\big)}{h}\\ &= \lim_{h \to 0} \frac{f(b + h\gamma(h) )-f(b)}{h} \end{align*}

Now for the sneaky bit. We can turn $f(b + h\gamma(h) )$ into $f(b+H)$ by setting

\begin{gather*} H = h\gamma(h) \end{gather*}

Now notice that as $h \to 0$ we have

\begin{align*} \lim_{h \to 0} H &= \lim_{h \to 0} h \cdot \gamma(h)\\ &= \lim_{h \to 0} h \cdot \lim_{h \to 0} \gamma(h)\\ &= 0 \cdot g'(a) = 0 \end{align*}

So as $h\to 0$ we also have $H \to 0\text{.}$

We now have

\begin{align*} F'(a) &= \lim_{h \to 0} \frac{f\big(b + H\big)-f(b)}{h}\\ &= \lim_{h \to 0} \underbrace{\frac{f\big(b + H\big)-f(b)}{H}}_{= \varphi(H) } \cdot \underbrace{\frac{H}{h}}_{ = \gamma(h)} & \text{if } H= h \gamma(h) \ne 0\\ &= \lim_{h \to 0}\big( \varphi(H) \cdot \gamma(h) \big)\\ &= \lim_{h \to 0} \varphi(H) \cdot \lim_{h \to 0} \gamma(h) & \text{since $H\to0$ as $h\to 0$}\\ &= \lim_{H \to 0} \varphi(H) \cdot \lim_{h \to 0} \gamma(h) &= f'(b) \cdot g'(a) \end{align*}

This is exactly the RHS of the chain rule. It is possible to have $H=0$ in the second line above. But that possibility is easy to deal with:

• If $g'(a)\ne 0\text{,}$ then, since $\lim_{h \to 0} \gamma(h) = g'(a)\text{,}$ $H= h \gamma(h)$ cannot be $0$ for small nonzero $h\text{.}$ Technically, there is an $h_0\gt 0$ such that $H= h \gamma(h)\ne 0$ for all $0 \lt |h| \lt h_0\text{.}$ In taking the limit $h\to 0\text{,}$ above, we need only consider $0 \lt |h| \lt h_0$ and so, in this case, the above computation is completely correct.
• If $g'(a)=0\text{,}$ the above computation is still fine provided we exclude all $h$'s for which $H= h \gamma(h)\ne 0\text{.}$ When $g'(a)=0\text{,}$ the right hand side, $f'\big(g(a)\big) \cdot g'(a)\text{,}$ of the chain rule is $0\text{.}$ So the above computation gives
\begin{equation*} \lim_{\genfrac{}{}{0pt}{}{h \to 0}{\gamma(h)\ne 0}} \frac{f\big(b + H\big)-f(b)}{h} =f'\big(g(a)\big) \cdot g'(a) = 0 \end{equation*}
On the other hand, when $H=0\text{,}$ we have $f\big(b + H\big)-f(b)=0\text{.}$ So
\begin{equation*} \lim_{\genfrac{}{}{0pt}{}{h \to 0}{\gamma(h) = 0}} \frac{f\big(b + H\big)-f(b)}{h} =0 \end{equation*}
too. That's all we need.