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Calculus 1 Problems & Solutions  –  Chapter 4  –  Section 4.2.2

4.2.2
The Natural Exponential Function

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Review

1. The exp Function

The graph of the natural logarithm function y = ln x is shown in Fig. 1.1. It's clear from the graph that ln is one-to-one.
Let's show it. Suppose ln x₁ = ln x₂. Then ln 1 = 0 = ln x₁ – ln x₂ = ln x₁/x₂. So x₁/x₂ = 1, thus x₁ = x₂. Indeed ln is
one-to-one. Consequently, by Section 2.3.4 Review Part 2, ln has an inverse function. For the moment let's call this

Fig. 1.1 y = ln x is one-to-one, so has an inverse function, for the moment called exp:

We have:

exp 0 = 1,

since if y = exp 0, then ln y = 0 = ln 1, hence, as ln is one-to-one, y = 1.

As ln and exp are inverse functions of each other, we have:

ln exp x = x,
exp ln x = x,

as pictured in Fig. 1.2.

Fig. 1.2 ln exp x = ln y = x, exp ln x = exp y = x.

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2. A Law Of Exponents

We've seen above that exp 0 = 1. The exp function satisfies the law of exponents a⁰ = 1. We now show that exp also
satisfies another law of exponents, namely (a^x)^r = a^rx.

Theorem 2.1

For any rational number r we have:

(exp x)^r = exp rx.

Proof
Let y = (exp x)^r. Then ln y = ln(exp x)^r = r ln exp x = rx. So (exp x)^r = y = exp rx.
EOP

For the meaning of a^r (a > 0) where r is rational, see Section 4.2.1 Review Part 4. We've not yet defined the meaning of
a^s where s is irrational. That's why we handle only the rationals r in this theorem. The meaning of a^s where s is irrational
will be dealt with in this section (e^s) and the next (a^s, any a > 0).

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3. The Number e

As 1 is in range(ln) = R, by the intermediate-value theorem (see Section 1.2.3 Theorem 1.1) there exists x = x₁ such
that ln x₁ = 1. Let's denote x₁ by e, so that ln e = 1; see Fig. 3.1. So e = exp 1. The number e can also be defined using
exp as follows: since 1 is in dom(exp) = R, exp 1 is defined; let e = exp 1. We prefer to define e using an equation of
the form e = some expression. Thus we use the exp function.

Definition 3.1

The number e is defined by:

e = exp 1.

Thus:

ln e = 1.

Fig. 3.1 ln e = 1, e = exp 1.

Fig. 3.2 Colored Area = ln e = 1 Square Unit.

In Section 4.2.1 we found that 2.70 < e < 2.75. In Remark 5.1 below we'll show that e defined in this section is the same
number as e defined in that section.

By Section 4.2.1 Definition 2.1, the area of the plane region bounded by graph of y = 1/t, the t-axis, the vertical line t = 1,
and the vertical line t = e is ln e = 1 square unit, as shown in Fig. 3.2.

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4. The Natural Exponential Function

By Theorem 2.1 we have exp rt = (exp t)^r for any rational number r. So, for t = 1:

exp r = exp 1r = (exp 1)^r = e^r

for any rational number r. Yes the value exp r where r is rational is equal to e raised to the power of r. The function
exp r on the rationals is the exponential function e^r on the rationals! We wish to extend this property to all the reals, to
get exp x = e^x for all real x. We thus have to define the meaning of e^s where s is irrational. Well, exp x is defined for all
real x, including the irrationals. Now, we've got the result that e^r = exp r for all rational r. Consequently it's only natural
to define e^s, where s is irrational, to be exp s.

For any irrational number s, e^s is defined by: e^s = exp s.

Hence e^x = exp x for all real x. We can now make the following definition, which is consistent with e^r for rationals r.

Definition 4.1

The natural exponential function is denoted by e^x and is defined by:

e^x = exp x

for all real number x. For simplicity this function is also called just the exponential function.

We now recognize that the exp function is the natural exponential function. That's why we denoted it “exp”. Hereafter we
drop the notation exp x in favor of the notation e^x. Since exp is the inverse of the natural logarithm, we now can state
the following:

The inverse of the natural logarithm function is the natural exponential function:

The inverse property [4.1] justifies the defining of the natural logarithm function as such – see the paragraph that follows
Section 4.2.1 Definition 2.1. Properties [4.2] and [4.3] can be proved as follows. Let y = e^lnx. It follows that ln y = ln x,
therefore y = x, ie e^lnx = x, and property [4.2] is proved. The case for property [4.3] is similar. Also see Fig. 1.2.

The Natural Logarithm And The Number e

Let y = e^x. So x = exponent on e (exponent to which e is raised) to get y. As y = e^x we have x = ln y. Thus the natural
logarithm of y is the exponent on e to get y. The natural logarithm of a positive number is the exponent on e to get that
number. The natural logarithm is the logarithm with base e.

Example 4.1

Simplify the following expressions.

Solution

`

EOS

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5. Differentiation Of The Natural Exponential Function

We now show that the derivative of e^x is e^x itself. This is perhaps the most important property of the natural exponential
function. Also see Section 4.2.1.

Theorem 5.1

For any real number x we have:

Proof
Let y = e^x. So x = ln y. Differentiating this equation implicitly with respect to x we get:

EOP

Note that (d/dx) e^x is not xe^x–1! We have (d/dx) xⁿ = nx^n–1, while (d/dx) e^x = e^x. The power function xⁿ has the variable
at the base, while the exponential function e^x has the variable at the exponent.

Remark 5.1

In Section 4.2.1 we defined e as the number such that:

ie the derivative of e^x at x = 0.

Example 5.1

Differentiate y = e^sin3x.

Solution

dy/dx = e^sin3x(d/dx) sin 3x = e^sin3xcos 3x (d/dx) 3x = 3e^sin3xcos 3x.

EOS

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6. Graph Of The Natural Exponential Function

The graph of the natural exponential function y = e^x is the mirror image of that of its inverse, the natural logarithm
function y = ln x, in the line y = x. It's sketched in Fig. 6.1. Note that:

Fig. 6.1 Graph of y = e^x is mirror image of that of y = ln x in line y = x.

Note the following:

Also,  y' = e^x > 0 for all x and y'' = e^x > 0 for all x, confirming that y = e^x is increasing and concave up on R.

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7. Properties Of The Natural Exponential Function

We now establish the familiar properties of the natural exponential function, properties associated with powers e^x of a
fixed base and a variable exponent. They're the laws of exponents.

Theorem 7.1 – The Laws Of Exponents

a. e^x+y = e^xe^y.

e. e⁰ = 1.

Proof
a. ln e^x+y = x + y = ln e^x + ln e^y = ln(e^xe^y), so that e^x+y = e^xe^y because ln is one-to-one.
b. ln e^–x = –x = –(ln e^x) = ln(1/e^x), so that e^–x = 1/e^x.
c. ln e^x–y = x – y = ln e^x – ln e^y = ln(e^x/e^y), so that e^x–y = e^x/e^y.
d. ln(e^x)^y = y ln e^x = xy = ln e^xy, so that (e^x)^y = e^xy.
e. e⁰ = e^1–1 = e¹/e¹ = 1.
EOP

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8. Representation Of e^x As A Limit

Let's compare e² to (1 + (2/n))ⁿ for some larger and larger values of the positive integer n. Employing our calculator we
have:

Theorem 8.1

For any real number x we have:

Proof

Since ln is differentiable, it's continuous^{8.1}. So we get^{8.2}:

EOP

^{8.1} See Section 2.2 Theorem 1.1.

^{8.2} See Section 1.2.1 Theorem 3.1 iv a.

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9. Approximations Of e

Setting x = 1 in Eq. 8.1 we get:

This formula can be used to compute approximations of e, as shown in Fig. 9.1.

Fig. 9.1 Computation Of Approximations Of e. The approximation gets better and better as n > 0 gets larger and larger.

For any integer n > 0 let n! = 1 x 2 x 3 x ... x n. The notation n! is read “n factorial”, for the reason that it's a product of
positive integer factors from 1 to n. For example, 3! = 1 x 2 x 3 = 6. Define 0! = 1. It can be shown that a better way to
compute approximations of e is to utilize the infinite series (sum of infinitely many terms):

which is usually encountered in a second calculus course or in a first analysis course.

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10. The Fastness Of e^x

Let's compare the sizes of e^x and x^1,000. Surely at x = 2 we have e² < 2^1,000, and at x = 3 we still have e³ < 3^1,000. Now
consider x = 10,000. With the help of our calculator we obtain:

Theorem 10.1

For any integer n > 0 we have:

Proof

EOP

{¹} For the last equation see Section 4.2.1 Problem & Solution 5.

Remarks 10.1

b. For a given integer n > 0, no matter how large it is, e^x may be smaller than xⁿ only for x in a finite range, say for x in
    (1, b), where b > 1 is finite. As x grows larger and larger away from b, e^x grows larger and larger than xⁿ does.

c. This “fastness” property of e^x has led to the popular “exponential” phrases such as “exponential growth” or “increase
    exponentially”. It's the inverse of the “slowness” property of ln x, which was demonstrated in Section 4.2.1 Problem &
Solution 5. That's consistent with the fact that e^x is the inverse function of ln x.

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Problems & Solutions

1.  Simplify the following expressions.

a.e^{3 ln 5}e^{–ln 25}.
b. e^{2 lncosx} + ln²e^sinx.

Solution

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2.  Find:

a.  (d/dx) (Ae^ax cos bx + Be^ax sin bx).

Solution

a.  (d/dx) (Ae^ax cos bx + Be^ax sin bx) = Aae^ax cos bx– Abe^ax sin bx + Bae^ax sin bx + Bbe^ax cos bx
= (Aa + Bb) e^ax cos bx + (Ba – Ab) e^ax sin bx.

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3.  Find the derivative of any order of y = xe^ax as follows:

a.  Find its first three derivatives.
b.  Guess a formula for its nth derivative, where n is any positive integer.
c.  Prove your guess using mathematical induction.

Solution

a. y' = e^ax + axe^ax,
y'' = ae^ax + ae^ax + a²xe^ax = 2ae^ax + a²xe^ax,
y''' = 2a²e^ax + a²e^ax + a³xe^ax = 3a²e^ax + a³xe^ax.

b. y⁽ⁿ⁾ = na^n–1e^ax + aⁿxe^ax.

c.  The formula is true for n = 1. Suppose it's true for an arbitrary positive integer n. Then we have:

y⁽ⁿ⁾ = na^n–1e^ax + aⁿxe^ax,

y⁽ⁿ⁺¹⁾ = (d/dx) y⁽ⁿ⁾
             = naⁿe^ax + aⁿe^ax + aⁿ⁺¹xe^ax
             = (n + 1)aⁿe^ax + aⁿ⁺¹xe^ax
             = (n + 1)a^(n+1)–1e^ax + aⁿ⁺¹xe^ax.

     So the formula is also true for n + 1. Thus it's true for all positive integers.

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4.  Find the equation of the straight-line tangent to the curve y = ln x and passing thru the origin. Sketch the curve and
     tangent.

Solution

Since (d/dx) ln x = 1/x, the equation of the tangent line at any point (x₁, y₁) = (x₁, ln x₁) is y = ln x₁ + (1/x₁)(x – x₁) or
y = x/x₁ + ln x₁ – 1. The tangent line will pass thru the origin if 0 = 0/x₁ + ln x₁ – 1, or ln x₁ = 1, or x₁ = e. Hence the
equation of that tangent line is y = x/e + ln e – 1, or y = x/e.

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5.  Find the slope of the curve:

Solution

Differentiating the given equation implicitly with respect to x we get:

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