Proof of Product and Quotient RulesDate: 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/ |
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