So far we've been successful with a divide-and-conquer approach to differentiation: the product rule and the chain rule offer methods of breaking a function down into simpler parts, and finding
the derivative of the whole thing based on knowledge of the derivatives of the parts. We know how to find the derivatives of sums, differences, and products, so the obvious next step is to look
for a way of handling division. This is straightforward, since we know that the derivative of the function
is 
.
Let
and 
be
functions of 
.
Then by the product rule,
and by the chain rule,
This is so easy to rederive on demand that I suggest not memorizing it.
By the way, notice how the notation becomes a little awkward when we want to write a derivative like 
. When we're differentiating a complicated function, it can be uncomfortable trying to cram the expression into the top of the
d.../d... fraction. Therefore it would be more common to write such an expression like this:
This could be considered an abuse of notation, making d look like a number being divided by another number 
, when actually 
is meaningless on its own. On the other hand, we can consider the symbol 
to represent the operation of differentiation with respect to 
; such an interpretation will seem more natural to those who have been inculcated with
the taboo against considering inffnitesimals as numbers in the first place.
Using the new notation, the quotient rule becomes
The interpretation of the minus sign is that if 
increases, 
decreases.
Example
Differentiate 
, and check that the result makes sense.
We identify 
with 
and 
with 
.
The result is
One way to check that the result makes sense is to consider extreme values of 
. For very large values of 
, the 1 on the bottom of
becomes negligible compared to the 
, and the function 
approaches 
as a limit. Therefore we expect that the derivative 
should approach zero, since the derivative of a constant is zero. It works: plugging in bigger and bigger numbers for
in the expression for the derivative
does give smaller and smaller results. (In the second term, the denominator gets bigger faster than the numerator, because it has a square in it.)
Another way to check the result is to verify that the units work out. Suppose arbitrarily that x has units of gallons. (If the 3 on the bottom is unitless, then the 1 would have to represent
1 gallon, since you can’t add things that have different units.) The function 
is defined by an expression with units of gallons divided by gallons, so 
is unitless. Therefore the derivative 
should have units of inverse gallons. Both terms in the expression for the
derivative do have those units, so the units of the answer check out.
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