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Section 2.3—The Product Rule
In this section, we will develop a rule for differentiating the product of two
functions, such as f 1x2 13x 2 121x 3 82 and g1x2 1x 32 3 1x 22 2 ,
without first expanding the expressions.
You might suspect that the derivative of a product of two functions is simply the
product of the separate derivatives. An example shows that this is not so.
EXAMPLE 1
Reasoning about the derivative of a product of two functions
Let p1x2 f 1x2g1x2, where f 1x2 1x 2 22 and g1x2 1x 52.
Show that p¿1x2 f ¿1x2g¿1x2.
Solution
The expression
p1x2
can be simplified.
p1x2 1x 2 221x 52
x 3 5x 2 2x 10
p¿1x2 3x 2 10x 2
f ¿1x2 2x and g¿1x2 1, so f ¿1x2g¿1x2 12x2112 2x.
Since 2x is not the derivative of p1x2, we have shown that p¿1x2 f ¿1x2g¿1x2.
The correct method for differentiating a product of two functions uses the
following rule.
The Product Rule
If p1x2 f 1x2g1x2, then p¿1x2 f ¿1x2g1x2 f 1x2g¿1x2.
If u and v are functions of x, d du
1uv2
dx dx v u dv
dx .
In words, the product rule says, “the derivative of the product of two functions is
equal to the derivative of the first function times the second function plus the first
function times the derivative of the second function.”
Proof:
p1x2 f 1x2g1x2, so using the definition of the derivative,
p¿1x2 lim
hS0
f 1x h2g1x h2 f 1x2g1x2
.
h
NEL CHAPTER 2 85