Differentiabilty and continuous difference quotients

Recall the definition of differentiable. $g:E\rightarrow(-\infty,\infty)$ is differentiable at $b\in E$ means that the limit as h approaches of

$$\frac{g(b+h) - g(b)}{h}$$

exists. We denote this limit by g'(b). Suppose that g is differentiable at b. Let us define a function DQ by the rule

$$DQ(h) = \left\{ \begin{array} {cc} \displaystyle{\frac{g(b+h... ...if}\;b+h\in E,\;h\neq 0\\ g'(b) & {\rm h = 0}\end{array}\right.$$

From the definitions of continuous and of differentiable, DQ is continous at h = 0. Let $f:E\rightarrow (-\infty,\infty)$ be given by f(x) = x-b. This function is continuous, its range is the domain of DQ, and $\lim_{x\rightarrow b}f(x) = 0$. Therefore

$$\lim_{x\rightarrow b} DQ(f(x)) = DQ(0) = g'(b).$$

Since for $x\neq b$ we have $f(x) \neq 0$ and

$$DQ(f(x)) = \frac{g(b + (x-b))-g(b)}{x-b}\;{\rm if}\; x\neq b,$$

we have shown that

$$\lim_{x\rightarrow b} \frac{g(x) - g(b)}{x-b} = g'(b).$$

In fact, we can do much better. If $c:D\rightarrow E$, c(a) = 0 and c is continuous at a, then

$$\lim_{x\rightarrow a} DQ(c(x)) = g'(b).$$

This is what we need to prove the chain rule. Suppose that f is differentiable at a and g is differentiable at f(a). Put c(h) = f(a+h)-f(a). Since f is differentiable at a we know that f is continuous at a, so c is continuous at h = 0. (What is the domain of c?) Observe that if b = f(a)

$$DQ(f(a+h)-f(a))\frac{f(a+h)-f(a)}{h} = \frac{g(f(a+h))-g(f(a))}{h}$$

since neither side is defined for h = 0 and for $h\neq 0$ if f(a+h) - f(a) = 0 we have

$$\begin{array} {rcl} {\displaystyle DQ(f(a+h)-f(a))\frac{f(a+... ...displaystyle \frac{g(f(a)) - g(f(a))}{h}}\\ & = & 0,\end{array}$$

while if $f(a+h)- f(a)\neq 0$ then

$$DQ(f(a+h)-f(a))\frac{f(a+h)-f(a)}{h} = \frac{g(f(a+h)) - g... ...f(a)}\times\frac{f(a+h)-f(a)}{h} = \frac{g(f(a+h))-g(f(a))}{h}$$

Therefore

$$\lim_{h\rightarrow 0}\frac{g(f(a+h)) - g(f(a))}{h} = \lim_... ...ghtarrow 0}DQ(f(a+h)-f(a))\frac{f(a+h)-f(a)}{h} =g'(f(a))f'(a).$$

We can also make the reverse argument. Suppose that

$$\lim_{x\rightarrow b} \frac{g(x)-g(b)}{x-b} = L$$

Then

$$L = \lim_{h\rightarrow 0} \frac{g(b+h)-g(b)}{b+h-b} = \lim_{h\rightarrow 0} \frac{g(b+h)-g(b)}{h} = g'(b)$$

Again, more generally, if $\lim_{x\rightarrow a} f(x) = f(a)$ and g is differentiable at f(a) then

$$\lim_{x\rightarrow a}{g(f(x))-g(f(a))}{f(x) - f(a)} = g'(f(a)).$$