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

Section 3.6 Derivatives of Logarithmic Functions 219
In general, if we combine Formula 2 with the Chain Rule as in Example 1, we get
3
d
dx sln ud − 1 u
du
dx
or
d
t9sxd
fln tsxdg −
dx tsxd
ExamplE 2 Find d lnssin xd.
dx
SOLUTION Using (3), we have
d
1
lnssin xd −
dx sin x
d
1
ssin xd − cos x − cot x
dx sin x ■
ExamplE 3 Differentiate f sxd − sln x .
SOLUTION This time the logarithm is the inner function, so the Chain Rule gives
f 9sxd − 1 2 sln xd21y2 d dx sln xd − 1
2sln x
? 1 x − 1
2xsln x
■
ExamplE 4 Differentiate f sxd − log 10 s2 1 sin xd.
SOLUTION Using Formula 1 with b − 10, we have
Figure 1 shows the graph of the
function f of Example 5 together with
the graph of its derivative. It gives a
visual check on our calculation. Notice
that f 9sxd is large negative when f is
rapidly decreasing.
y
1
0
fª
FIGURE 1
f
x
f 9sxd − d dx log 10s2 1 sin xd
−
−
ExamplE 5 Find d dx ln x 1 1
sx 2 2 .
SOLUTION 1
d
dx ln x 1 1
−
sx 2 2
1
s2 1 sin xd ln 10
cos x
s2 1 sin xd ln 10
1
x 1 1
sx 2 2
− sx 2 2
x 1 1
d
dx
x 1 1
sx 2 2
− x 2 2 2 1 2 sx 1 1d
sx 1 1dsx 2 2d
−
x 2 5
2sx 1 1dsx 2 2d
d
s2 1 sin xd
dx
sx 2 2 ∙ 1 2 sx 1 1d( 1 2)sx 2 2d 21y2
x 2 2
■
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