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

Section 3.2 The Product and Quotient Rules 187
ExamplE 5 Find an equation of the tangent line to the curve y − e x ys1 1 x 2 d at
the point s1, 1 2 ed.
SOLUtion According to the Quotient Rule, we have
s1 1 x 2 d d dy
dx − dx se x d 2 e x
s1 1 x 2 d 2
d
dx s1 1 x 2 d
− s1 1 x 2 de x 2 e x s2xd
s1 1 x 2 d 2
− e x s1 2 2x 1 x 2 d
s1 1 x 2 d 2
2.5
y= ´
1+≈
1
y= e
2
− e x s1 2 xd 2
s1 1 x 2 d 2
So the slope of the tangent line at s1, 1 2ed is
dy
dx
Z − 0
x−1
_2 3.5
0
FIGURE 4
This means that the tangent line at s1, 1 ed 2 is horizontal and its equation is y − 1 2e. fSee
Figure 4. Notice that the function is increasing and crosses its tangent line at s1, 1 2 ed.g ■
NOTE Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s
easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation.
For instance, although it is possible to differentiate the function
Fsxd − 3x 2 1 2sx
x
using the Quotient Rule, it is much easier to perform the division first and write the function
as
Fsxd − 3x 1 2x 21y2
before differentiating.
We summarize the differentiation formulas we have learned so far as follows.
Table of Differentiation Formulas
d
dx scd − 0
d
dx sx n d − nx n21
d
dx se x d − e x
scf d9 − cf9 s f 1 td9 − f91 t9 s f 2 td9 − f92 t9
s ftd9 − ft9 1 tf9
S f tD 9 −
tf9 2 ft9
t 2
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