2.6. QuotientsSo far we've been successful with a divideandconquer 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 1/u=u^{1} is u^{2}. Let u and v be functions of x. Then by the product rule, \[ \frac{d(v/u)}{dx} = \frac{dv}{dx} \cdot \frac{1}{u} + v \cdot \frac{d(1/u)}{dx} \] and by the chain rule, \[ \frac{d(v/u)}{dx} = \frac{dv}{dx} \cdot \frac{1}{u}  v \cdot \frac{1}{u^2}\frac{du}{dx}\] 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 d(v/u)/dx. 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: \[\frac{d}{dx} \left(\frac{v}{u}\right)\]
This could be considered an abuse of notation, making d look like a number being divided by another number dx, when actually d is meaningless on its own. On the other hand, we can consider the symbol d/dx to represent the operation of differentiation with respect to x; such an interpretation will seem more natural to those who have been inculcated with the taboo against considering infinitesimals as numbers in the first place. Using the new notation, the quotient rule becomes \[\frac{d}{dx} \left(\frac{v}{u}\right) = \frac{1}{u}\cdot\frac{dv}{dx}  \frac{v}{u^2} \cdot \frac{du}{dx} .\] The interpretation of the minus sign is that if u increases, v/u decreases. Example 25◊ Differentiate y = x/(1+3x), and check that the result makes sense. ◊ We identify v with x and u with 1+x. The result is \begin{align} \frac{d}{dx} \left(\frac{v}{u}\right) &= \frac{1}{u}\cdot\frac{dv}{dx}  \frac{v}{u^2} \cdot \frac{du}{dx} \\ &= \frac{1}{1+3x}  \frac{3x}{(1+3x)^2} \end{align} One way to check that the result makes sense is to consider extreme values of x. For very large values of x, the 1 on the bottom of x/(1+3x) becomes negligible compared to the 3x, and the function y approaches x/3x=1/3 as a limit. Therefore we expect that the derivative dy/dx should approach zero, since the derivative of a constant is zero. It works: plugging in bigger and bigger numbers for x 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 y is defined by an expression with units of gallons divided by gallons, so y is unitless. Therefore the derivative dy/dx 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.
