Proof of Product and Quotient Rules

Date: 02/07/2002 at 21:30:53
From: Theron Schaub
Subject: The proof of the product and quotient rules

I would like to know how the product rule and the quotient rule came 
about so I can better understand calculus. I have been to the math 
tutors and they don't know. 

Thank you, 
Theron

Date: 02/07/2002 at 22:59:13
From: Doctor Peterson
Subject: Re: The proof of the product and quotient rules

Hi, Theron.

Let's try using the definition of the derivative, and see what 
happens:

Suppose we have two differentiable functions f and g, and we have 
defined a new function p(x) = f(x) * g(x). Its derivative will be

            p(x+h) - p(x)       f(x+h)g(x+h) - f(x)g(x)
p'(x) = lim ------------- = lim -----------------------
       h->0       h        h->0            h

I'd like to get this to include something that looks like the 
definition of f'; I'll try adding and subtracting an intermediate 
term:
                          _______________________
                         /                       \
            f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)
      = lim -------------------------------------------------
       h->0                         h

That lets me factor something out of each pair:

            [f(x+h) - f(x)]g(x+h) - f(x)[g(x+h) - g(x)]
      = lim -------------------------------------------
       h->0                     h

             f(x+h) - f(x)               g(x+h) - g(x)
      = lim [------------- g(x+h) - f(x) -------------]
       h->0        h                           h

A little magic with the limits (which has to be proven valid), and we 
get

            f(x+h) - f(x)                               g(x+h) - g(x)
      = lim ------------- * lim g(x+h) - lim f(x) * lim -------------
                  h                                           h

      = f'(x) g(x) - f(x) g'(x)

We've got it!

Now let's look at what this means. Suppose the length of a rectangle 
is varying with time according to a function f(t), and the width is 
g(t). During a small time interval from t to t+dt, how has the area 
f(t)g(t) changed?

           g(t)    g'(t) dt
    +--------------+---+
    |              |   |
    |              |   |
    |              |   |f(t)
    |              |   |
    |              |   |
    +--------------+---+
    |              |   |f'(t) dt
    +--------------+---+

As shown, we can approximate the change in f(t) by f'(t) dt, and the 
change in g(t) by g'(t) dt. The change in the area fg consists of the 
two long rectangles and the little square:

    fg'dt + f'gdt + f'g'dt^2

Divide this by dt, and we have

    fg' + f'g + f'g'dt

Since dt is very small, we can ignore the conribution from the tiny 
square, and the derivative is fg' + f'g. That's not a formal proof, 
but it gives a feel for what is happening. Stare at the picture for a 
while, and you'll see why each derivative is multiplied by the other 
function, and why there are two terms added together.

    +--------------+---+
    |              |   |
    |              |   |
    |      fg      |f'g|
    |              |   |
    |              |   |
    +--------------+---+
    |      fg'     |
    +--------------+

To get the quotient rule, just apply the product rule to

    q(x) = f(x) [1/g(x)]

using the chain rule to find the derivative of 1/g(x) = g(x)^-1.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/