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

222 Chapter 3 Differentiation Rules
Figure 3 illustrates Example 8 by
showing the graphs of f sxd − x sx and
its derivative.
y
1
0
FIGURE 3
1
fª
f
x
ExamplE 8 Differentiate y − x sx .
SOLUTION 1 Since both the base and the exponent are variable, we use logarithmic
differentiation:
ln y − ln x sx − sx ln x
y9
y − sx ? 1 x 1 sln xd 1
2sx
y9 − yS 1
sx
1 ln x
2sx
D − x sx S
SOLUTION 2 Another method is to write x sx − se ln x d sx :
The Number e as a Limit
D
2 1 ln x
2sx
d
dx sxsx d − d dx sesx ln x d − esx ln x d dx ssx ln xd
− x sx S
2 1 ln x
2sx
D (as in Solution 1) ■
We have shown that if f sxd − ln x, then f 9sxd − 1yx. Thus f 9s1d − 1. We now use this
fact to express the number e as a limit.
From the definition of a derivative as a limit, we have
FIGURE 4
y
2
1
0
3
y=(1+x)!?®
x s1 1 xd 1yx
0.1 2.59374246
0.01 2.70481383
0.001 2.71692393
0.0001 2.71814593
0.00001 2.71826824
0.000001 2.71828047
0.0000001 2.71828169
0.00000001 2.71828181
x
f 9s1d − lim
h l 0
Because f 9s1d − 1, we have
f s1 1 hd 2 f s1d
h
− lim
x l 0
f s1 1 xd 2 f s1d
x
lns1 1 xd 2 ln 1 1
− lim
− lim lns1 1 xd
x l 0 x
x l 0 x
− lim
x l 0
lns1 1 xd 1yx
lim lns1 1
x l 0 xd1yx − 1
Then, by Theorem 2.5.8 and the continuity of the exponential function, we have
e − e 1 − e lim x l 0 lns11xd1yx − lim − lim s1 1
x l 0 elns11xd1yx xd1yx
x l 0
5 e − lim
x l 0
s1 1 xd 1yx
Formula 5 is illustrated by the graph of the function y − s1 1 xd 1yx in Figure 4 and a
table of values for small values of x. This illustrates the fact that, correct to seven decimal
places,
e < 2.7182818
If we put n − 1yx in Formula 5, then n l ` as x l 0 1 and so an alternative expression
for e is
6 e − lim 1 1 n
n l `S1
nD
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