On-line Math 21
On-line Math 21
2.2 Differentiation Rules
Theorem 5
The derivative of a product fg ,
is given by
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:
|
|
|
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,
|
|
| |
|
|
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:
|
|
| |
|
|
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
|
|
| |
|
|
|
Copyright (c) 2000 by David L. Johnson.
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On 27 Oct 2000, 01:06.