Differentiating quotients

Since every quotient (a/b) can be rewritten at $a\cdot(1/b)$ we will look first at differentiating reciprocals. Suppose that $g:D\rightarrow(-\infty,0)\cup (0,\infty)$ and g'(a) exists. Here is how to compute (1/g)'(a):

$$\begin{array} {rcl} (1/g)'(a) & = & \lim_{h\rightarrow 0}\f... ...ft(\frac{-1}{g(a)g(a+h)}\frac{g(a+h)-g(a)}{h}\right)\end{array}$$

We recognize that in this last expression the second fraction has a limit of g'(a) as h approaches . The sticky point is what happens to g(a+h) as h approaches . We suspect it has a limit of g(a). We will show this below. Taking this as known, we see that

$$(1/g)'(a) = \frac{-1}{g(a)^2}g'(a)$$

and the general quotient rule follows quite easily: Suppose that $f:D\rightarrow(-\infty,\infty)$, $g:D\rightarrow(-\infty,0)\cup (0,\infty)$and f'(a) and g'(a) exist. Here is how to compute (f/g)'(a):

$$\begin{array} {rcl} (f/g)'(a) & = & (f\cdot (1/g))'(a)\\ & ... ...right)\\ & = & \frac{f'(a)g(a) - f(a)g'(a)}{g(a)^2}\end{array}$$

There is no good way to try to remember this formula. We can observe that the negative sign with the g' makes sense in so far that if g increases, g' is positive, and as g increases, the fraction f/g) decreases, which should make the rate of change negative. The best advice is to practice, practice practice.

Here is an example: $h:(-\infty,\infty)\rightarrow(-\infty,\infty)$, $h(x) = \arctan(x)/(1+x^2)$. Then

$$\begin{array} {rcl} h'(x) & = & \frac{\arctan'(x)(1+x^2) - ... ...1+x^2)^2}\\ & = & \frac{1 - 2x\arctan(x)}{(1+x^2)^2}\end{array}$$

This idea can be used to differentiate $\cot(x) = \cos(x)/\sin(x)$, $\sec(x) = 1/\cos(x)$ and e^-x = 1/e^x.

Here is a fact that we can easily prove:
Theorem: If g'(a) exists then g is continuous at a.
Proof: Observe that

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

The fraction on the right has a limit of g'(a) as h approaches , so

$$\begin{array} {rcl} \lim_{h\rightarrow 0}g(a+h) & = & \lim_{... ...-g(a)}{h}\right)\\ & = & g(a) + 0g'(a)\\ & = & g(a),\end{array}$$

which is precisely what we mean when we say that g is continuous at a. QED