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

186 Chapter 3 Differentiation Rules
by amounts Dx, Du, and Dv, then the corresponding change in the quotient uyv is
DS u vD − u 1 Du
v 1 Dv 2 u v
−
su 1 Dudv 2 usv 1 Dvd
vsv 1 Dvd
−
vDu 2 uDv
vsv 1 Dvd
so
d
dxS
vD u − lim
Dx l 0
Dsuyvd
Dx
− lim
Dx l 0
v Du
Dx 2 u Dv
Dx
vsv 1 Dvd
As Dx l 0, Dv l 0 also, because v − tsxd is differentiable and therefore continuous.
Thus, using the Limit Laws, we get
d
dxS
vD u −
v lim
Dx l 0
Du
Dx 2 u lim
Dx l 0
−
v lim sv 1 Dvd
Dx l 0
Dv
Dx
v du
dx 2 u dv
dx
v 2
In prime notation:
StD f 9 tf 9 2 ft9
−
t 2
The Quotient Rule If f and t are differentiable, then
d
F f tsxd
−
dx tsxdG d dx f f sxdg 2 f sxd d dx ftsxdg
ftsxdg 2
We can use a graphing device to
check that the answer to Example 4
is plausible. Figure 3 shows the graphs
of the function of Example 4 and its
derivative. Notice that when y grows
rapidly (near 22), y9 is large. And
when y grows slowly, y9 is near 0.
1.5
_4 4
y
yª
In words, the Quotient Rule says that the derivative of a quotient is the denominator
times the derivative of the numerator minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.
The Quotient Rule and the other differentiation formulas enable us to compute the
derivative of any rational function, as the next example illustrates.
ExamplE 4 Let y − x 2 1 x 2 2
. Then
x 3 1 6
sx 3 1 6d d dx sx 2 1 x 2 2d 2 sx 2 1 x 2 2d d dx sx 3 1 6d
y9 −
sx 3 1 6d 2
− sx 3 1 6ds2x 1 1d 2 sx 2 1 x 2 2ds3x 2 d
sx 3 1 6d 2
− s2x 4 1 x 3 1 12x 1 6d 2 s3x 4 1 3x 3 2 6x 2 d
sx 3 1 6d 2
FIGURE 3
_1.5
− 2x 4 2 2x 3 1 6x 2 1 12x 1 6
sx 3 1 6d 2
■
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