On-line Math 21

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.


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On 27 Oct 2000, 01:06.