James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

184 Chapter 3 Differentiation Rules
Recall that in Leibniz notation the
definition of a derivative can be written
as
dy
dx − lim
Dx l 0
Dy
Dx
If we now let Dx l 0, we get the derivative of uv:
d
Dsuvd
suvd − lim − lim Su Dv
dx Dx l 0 Dx Dx l 0 Dx 1 v Du Dv
1 Du
Dx DxD
− u lim
Dx l 0
Dv
Dx 1 v lim Du
S
Dx l 0 Dx 1 lim DS Du lim
Dx l 0 Dx l 0
− u dv
dx 1 v du
dx 1 0 ? dv
dx
Dv
DxD
2
d dv
suvd − u
dx dx 1 v du
dx
(Notice that Du l 0 as Dx l 0 since f is differentiable and therefore continuous.)
Although we started by assuming (for the geometric interpretation) that all the quantities
are positive, we notice that Equation 1 is always true. (The algebra is valid whether
u, v, Du, and Dv are positive or negative.) So we have proved Equation 2, known as the
Product Rule, for all differentiable functions u and v.
In prime notation:
s ftd9 − ft9 1 tf 9
The Product Rule If f and t are both differentiable, then
d
dx f f sxdtsxdg − f sxd d dx ftsxdg 1 tsxd d dx f f sxdg
In words, the Product Rule says that the derivative of a product of two functions is the
first function times the derivative of the second function plus the second function times
the derivative of the first function.
ExamplE 1
(a) If f sxd − xe x , find f 9sxd.
(b) Find the nth derivative, f snd sxd.
Figure 2 shows the graphs of the
function f of Example 1 and its
derivative f 9. Notice that f 9sxd is
positive when f is increasing and
negative when f is decreasing.
fª f
_3 1.5
3
SOLUTION
(a) By the Product Rule, we have
f 9sxd − d dx sxe x d
− x d dx se x d 1 e x d dx sxd
− xe x 1 e x ∙ 1 − sx 1 1de x
(b) Using the Product Rule a second time, we get
f 99sxd − d dx fsx 1 1dex g
FIGURE 2
_1
− sx 1 1d d dx se x d 1 e d x sx 1 1d
dx
− sx 1 1de x 1 e x ? 1 − sx 1 2de x
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