On-line Math 21

2.2  Differentiation Rules

Theorem 5 The derivative of a product fg , is given by

(f·g)¢ = f¢·g+f·g¢.

Proof:

We have to do this by the definition. As we always start off,

(f·g)¢(x) = lim h® 0  (f·g)(x+h)-(f·g)(x) h = lim h® 0  f(x+h)g(x+h)-f(x)g(x) h .

Now comes the trick. We need to break the difference quotient up into pieces, to see the derivatives of f and g separately. The trick is to subtract and add f(x)g(x+h) , so that:

(f·g)¢(x) = lim h® 0  f(x+h)g(x+h)-f(x)g(x) h = lim h® 0  f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x) h .

Now we break this mess up into two different fractions,

(f·g)¢(x) = ¼ = lim h® 0  f(x+h)g(x+h)-f(x)g(x+h) h + lim h® 0  f(x)g(x+h)-f(x)g(x) h .

and we factor out, in the first limit, the coommon factor of g(x+h) and in the second limit we factor out the factor of f(x) . After that, the derivatives of f and g appear, and we can take those limits since we assume we know what the derivatives of f and g are:

(f·g)¢(x) = ¼ = lim h® 0  f(x+h)-f(x) h g(x+h)+ lim h® 0  f(x) g(x+h)-g(x) h = lim h® 0  f(x+h)-f(x) h lim h® 0  g(x+h)+f(x) lim h® 0  g(x+h)-g(x) h = f¢(x)g(x)+f(x)g¢(x).

Copyright (c) 2000 by David L. Johnson.