dy
––
dx
dy
––
dx
1
–
y
1
–
x
d
––
dx
e153
The curve y = 1/x
1
–
n
1
–
n
1
–
n
n
––
n+ 1
n
By the
FUNDAMENTAL THEOREM OF CALCULUS
, the
derivative of an area function is the original function
and so we have:
ln(x) =
Now consider the corresponding exponential func-
tion y= ex. By taking logarithms, we obtain ln(y) = x.
Differentiating yields = 1, and so = y= ex. This
establishes definition 3 stating that f(x) = exis the func-
tion that equals its own derivative.
Now that we know the derivative of y= ex, we can
compute its Taylor series. We obtain:
Setting x= 1 establishes definition 4.
It remains now to establish definition 1. From the
graph of the curve y= 1/x, it is clear that the region
between x= 1 and x= 1 + is sandwiched between a
rectangle of area and a rectangle of
area = ×1 = . Thus:
Multiplying through by nyields:
As nbecomes large, the quantity approaches
the value 1. It must be the case, then, that
approaches a value for which its logarithm is 1.
Consequently equals 1. This gives:
With regard to the issue of compound interest, it
is necessary to compute the limit .
limn
n
r
n
→∞ +
1
limn
n
ne
→∞ +
=11
lim ln
n
n
n
→∞ +
11
11
+
n
n
nn
n
+≤+
≤
1111ln
1
1111
nnn+≤+
≤ln
11
11
1
1n
n
n
×
+
=+
ex
xx
x=++ + +123
23
!!
L