Ln(1+ x - users.math.msu.edu

[Pages:18]6.4

1

y = x - ln(1 + x)

2

y=x

1

-1 -1

y = ln(1 + x)

-2

1

2

Figure 1: A few functions referenced in Proposition 1.

We begin with a few important inequalities. Proposition 1. (a) ln(1 + x) x for all x > -1. (b) 1 + x ex for all x R.

Proof. To prove (a), let f (x) = 1 - 1/x. Then f (x) 0, for x 1 and so,

^x

^x

0 f (t) dt =

1- 1

dt = x - 1 - ln x

1

1

t

Rearranging yields

ln x x - 1, for x 1

which is equivalent to ln(1 + x) x, for x 0. We leave it as an exercise to show (a) when -1 < x < 0.

To prove (b), we note that the exponential function is increasing on R. Now exponentiate both sides of (a) and we have shown (b), provided x > -1.

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2

To complete that proof, notice that x -1 implies that 1 + x 0. Now since ex > 0 for all real numbers, we have shown that (b) holds for all

x R.

Example 1. Relative Rates of Growth and Decay

Prove the following limit. (1)

lim ln x = 0 x x

Proof. We may assume that x > 1 throughout the proof. Now we claim

that

(2)

0 < ln x < 2 x

If the claim is true, then

(3)

0 < ln x < 2x = 2

xx

x

since x is positive.

Now the right-hand side of (3) approaches 0 as x approaches , hence (1) follows by the Squeeze Law.

To prove the claim we note that the first inequality below follows since x > 1, the second inequality follows because the logarithm is an increasing function and the last inequality follows by Proposition 1(a).

0 < ln x < ln(1 + x) x

Thus

0 < ln x < x

Using the power rule for logarithms and rearranging yields (2).

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3

6.4 General Exponential and Logarithm Functions

Properties of ax

In section 6.2 we saw that if a > 0 then a has a logarithm. We start with the following definition (which we saw last time).

Definition. If a > 0 then we define ax by

(4)

ax = ex ln a

We can now establish some properties for the general exponential function, ax.

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4

Theorem 2. Properties of Exponents for ax

For a > 0, b > 0 and x, y R we have

(a) ax ? ay = ax+y (b) (ax)y = axy (c) ax ? bx = (ab)x

Proof. (a) ax ? ay = ex ln a ? ey ln a = e(x+y) ln a = ax+y (b) (ax)y = ey ln ax = exy ln a = axy (c) ax ? bx = ex ln a ? ex ln b = ex(ln a+ln b) = ex ln ab = (ab)x

Remark. Using property (a) above it is easy to establish,

ax ay

=

ax-y

and

a-x

=

1 ax

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5

The Derivative of au

If a > 0, then (5)

d ax = d ex ln a dx dx

= ex ln a d (x ln a) dx

= ax ln a

This leads to the following general derivative formula.

If a > 0 and u is a differentiable function of x, then

(6)

d au = au ln a du

dx

dx

Notice that if a = e then ln a = ln e = 1 so that (6) simplifies to

d eu = eu ln e du = eu du

dx

dx dx

as we saw last time. As the author points out, this is the reason that mathematicians prefer working with base e.

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6

The chain rule works as expected. Example 2.

(a) d 53x+1 = 53x+1 (ln 5) (3) dx

(b)

d 2 tan x2 = 2 tan x2 (ln 2) dx

d tan x2 dx

= 2 tan x2 (ln 2) sec2 x2 (2x)

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7

We are now in position to compute the derivatives of expressions of the form f (x)g(x) in a uniform way. We begin with a simplified example.

Example 3. Find the derivatives of each of x4 and 5x by taking logarithms, differentiating, and solving for dy/dx. Note: This is equivalent to (5).

a) First let y = x4 and take the logarithm of both sides to obtain

ln y = 4 ln x

Differentiating both sides with respect to x produces

1 dy = 4 y dx x

Rearranging yields

dy dx

=

y

4 x

=

4x4 x

=

4x3

Of course it is much easier to use the power rule!

b) First let y = 5x and take the logarithm of both sides to obtain

ln y = x ln 5

Differentiating both sides with respect to x produces

1 y

dy dx

=

ln

5

Rearranging yields

dy = y ln 5 = 5x ln 5 dx

as we saw in (5).

The expressions in Example 3 had either a constant exponent or a constant base. How do we handle f (x)g(x) when both the base and the

exponent are nonconstant functions of x? Consider the next example.

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8

5

y=x x

123

Figure

2:

The

graph

of

y

=

xx

Example 4.

If x > 0 then f (x) = x x is a differentiable function (see Fig. 2) defined

by

(7)

f (x)

=

xx

=

ex

ln x

To compute its derivative, we appeal to logarithmic differentiation (or we could apply the chain rule to the right-hand side of (7)). So let y = x x and proceed as we did in Example 3.

ln y = x ln x

Differentiating produces

1 y

dy dx

=

2lnxx

+

x x

Rearranging and simplifying yields

dy = y dx

2lnxx

+

x x

= 2xxx(ln x + 2)

In other words, (8)

f (x)

=

xx 2x

(ln

x

+

2)

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