Calculus Of One Real Variable

7.1 
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1. The Exponential Functions 
Note that the exponential function y
= b^{x}
is different from the power function y = x^{b}. For example,
for b = 2 and x = 3, we
have x^{b}
= 3^{2} = 9 and b^{x}
= 2^{3} = 8. In the power function
x^{b}, the base x is variable and the exponent b is constant, while in
the exponential function b^{x},
the base b is constant and the
exponent x is variable. Remark that b^{x} is called an exponential
function because the variable is in the exponent. Observe that the
variable x is the exponent and the
value b^{x}
is the
exponential.
Definition
We know since high school what the exponential b^{x}
means if x is a rational number. A
rational number is a number that is of
or can be written in the form of the ratio or fraction m/n, where m
and n are integers and n > 0. All the integers are rational,
since every integer k can
be written as k/1. Recall that the
definition of b^{m}^{/}^{n}
is:
b^{3.1}, b^{3.14}, b^{3.141}, b^{3.1415}, b^{3.14159}, ...,
Definition 1.1 – The Exponential Function
The exponential function b^{x}, where the base b is a positive constant different from 1, is defined for all real x as follows: Thus the domain of b^{x} is all x and its range is all y > 0. 
Remark 1.1
Since b
> 0 and b^{–}^{x} = 1/b^{x} for any x > 0, we have
that b^{x} > 0 for any x, positive or 0
or negative. For example, 2^{–3} = 1/2^{3} =
1/8 > 0. That is, the domain of b^{x} is all x and its range is all y > 0.
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2. The Number e 
Differentiability
Let's differentiate the exponential function b^{x}. Note that so
far we don't yet know what (d/dx) b^{x}
is. And we cannot claim that
it's xb^{x}^{–1}! Remember, b^{x} is not a power function. We must go back to the
definition of the derivative. Let f (x) = b^{x}.
First let's
differentiate it at x =
0:
exists. Of course this limit is the derivative of b^{x}
at x = 0, and thus it's the
slope of the tangent to the curve y = b^{x}
at its
yintercept (0, 1), as b^{0}
= 1. It's of the indeterminate form 0/0. If it exists, its value of course
depends on b.
Introducing e
Clearly things would simplify a lot if the above limit exists
and equals 1, because then the derivative of the function is the
function itself, a property that's simple plus interesting. So, if there's a
value of b, called e, such that:
That is, b^{x} is a differentiable
function. This is an additional interesting and important property: the
existence of e assures that
b^{x}
is differentiable. So our task of showing that b^{x} is differentiable
reduces to showing that e exists.
We now summarize the interesting and amazing properties of e.
Properties Of e
If there exists a positive number called e such that: 
Conversely, let's show that (4) implies (2): if (4) is
true, then there's a number e
such that (d/dx)
e^{x}
= e^{x},
then (d/dx)
e^{x}_{x}_{=0} =
e^{x}_{x}_{=0} =
e^{0}
= 1, then (2) is true, so (4) implies (2). Thus properties (2) and (4) are
equivalent. As a consequence, we can use
(4) to search for our desired e,
the number such that the derivative of the function e^{x} is the
function e^{x} itself.
But just one note before we proceed to search for e. If f (x) is a nonexponential function, is it
necessary that the properties
(d/dx)
f (x)_{x}_{=0} = 1 and (d/dx) f (x) = f (x) are equivalent? Let's examine some
examples. Let f (x) = x +
2. Then (d/dx)
f (x)
= 1 for all x, in particular (d/dx) f (x)_{x}_{=0} = 1. For this f
(x), the properties aren't
equivalent. Another example: let f (x) =
sin
x. Then (d/dx) f (x) = cos x
and (d/dx)
f (x)_{x}_{=0} = cos 0 = 1. For this f (x),
the properties aren't equivalent either. So the
answer is no. For nonexponential functions, the properties aren't necessarily
equivalent.
A Range In which e Lies If It Exists
To get a feeling of whether or not e
may exist, let's use its limit property to determine a range in which it lies.
Let's give h a
small value fairly close to 0, say h =
0.001, and find a value of b
that makes the ratio (b^{h}
– 1)/h close to 1. We reach for a
Fig. 2.1 Finding value of b that makes ratio (b^{0.001} – 1)/0.001 close to 1. 
calculator and proceed as shown in Fig. 2.1. Clearly e most likely exists and the range in which it lies is 2.70 < e < 2.72.
Search For e
As its derivative e^{x} > 0 for all x, the function e^{x} is strictly
increasing, so is onetoone, thus has an inverse function. Let L(x) be
this inverse function. Let y = L(x),
so that x = e^{y}. We have:
Consequently, if e
exists, then there's a function L(x) that's the inverse of e^{x}
and that's an antiderivative of 1/x.
This gives us
an idea: find an antiderivative L(x) of 1/x,
then define a number called e
that will (1) lead us to establish that L(x) is the
inverse of e^{x},
and (2) be the desired e,
the number we search for, ie, such that (d/dx) e^{x} = e^{x}.
Now let's proceed to search for e,
using the idea described above. The process won't assume that e exists; of course it can't,
because it tries to show that e
exists. Instead, after finding an antiderivative of 1/x,
it'll define a number that exists by a
theorem. That number will be our desired e,
and hence will be called e
from the beginning. Then we'll have shown that our
desired e indeed exists.
Do We Already Have An Antiderivative Of 1/x?
Before we proceed to find an antiderivative of 1/x, let's ask ourselves if we already have
a function that's an antiderivative of
1/x. Since 1/x = x^{–1}, let's examine the integerpower
functions, as displayed in the table in Fig. 2.2. In the f
(x) column, every
Fig. 2.2 Examining IntegerPower Functions. 
integer power is present somewhere. However in the f '(x) column, the power x^{–1}
= 1/x is “annihilated” by 0. So
1/x is absent.
Now we accept that so far we don't yet have an antiderivative of 1/x.
Finding An Antiderivative Of 1/x
As we want to find an antiderivative of 1/x, we of course have to consider the
function 1/x. So let's consider the
function y =
1/x for x
> 0, whose graph is shown in Fig. 2.3. The reason why we use x > 0 only is that the antiderivative
of 1/x that we
want will have to be the inverse function of e^{x}, whose range is all y > 0, which implies that the domain
of that antiderivative is
all x > 0. For any u > 0, let A be the area of the plane region bounded
by the graph of y = 1/x, the xaxis,
the vertical line x =
1, and the vertical line x = u. The region is colored in Fig. 2.3.
Obviously, as u varies (while
remaining > 0), A
also varies.
That is, A
is a function of u. Thus it's only
natural to denote it A(u). Now we want to denote the variable of
the function A
by

Fig. 2.3 Colored area A(u) is a function of u. 

Fig. 2.4 

Fig. 2.5 L(x) = –A(x) if 0 < x < 1. 
the letter x,
instead of u, to conform to tradition.
In Fig. 2.4 we replace the letter x by t and the letter u
by x, to get the graph
of the function y = 1/t and the function A(x).
Note that the horizontal axis is renamed as the taxis,
and that instead of the
letter t, we can utilize any other
letter that's normally utilized for variables, like u
or v, as long as it's not the
alreadyused x or
y.
However it’s a common practice to utilize the letter t.
Recall that areas are positiveor0 quantities. Evidently,
A(x)
> 0 if x > 1 or 0 < x < 1 and A(1)
= 0. Define the function L(x)
as follows:
Clearly:
A_{1} < A < A_{2},

Fig. 2.6 (d/dx) L(x) = 1/x. 
By the Squeeze Theorem in Section 1.1.2 Theorem 5.1 applied to onesided limits we get:
Conclusion. It
follows that, as the right and lefthand limits are equal to 1/x, we obtain:
We've found our function: an antiderivative of 1/x is that function L(x).
Defining e
We want to define e
utilizing the function L(x) in such a way that the inverse of L(x)
will be e^{x}.
If y = L(x), then we should
have x = e^{y}. Let k = L(e). Then e
= e^{k}.
So k = 1. Thus we'll define e to be the number such that L(e) =
1.
We know that L(1) = 0. As seen in Fig. 2.7, it's clear that:
Now L(x) is differentiable for all x > 0, so it's continuous for all x > 0. As L'(x) = 1/x
> 0 for all x > 0, L(x) is
increasing. So
L(1) < 1 < L(4).
Thus, by the IntermediateValue Theorem, there exists a number, let's call it e, where 1 < e
< 4, such that
L(e) =
1. See Fig. 2.8. That is, there exists a number e,
where 1 < e < 4, such that the area
of the region bounded by y =
1/x,
the xaxis, the vertical line x = 1, and the vertical line x = e
equals 1. The region is colored in Fig. 2.8.
Fig. 2.7 L(4) > 1. 
Fig. 2.8 L(e) = 1. 
It's Our Desired e
We're now going to show that the e
as defined above is our desired e.
Intuitively, as evidenced in Fig. 2.8, as x
increases
from the nearright of 0 toward infinity, L(x) is strictly increasing. Formally, L'(x) = 1/x
> 0 for all x > 0, so L(x) is
a strictly
increasing function. Thus it's onetoone, and consequently it has an inverse
function. Let E(x)
be this inverse function.
First we show that E(x) = e^{x}. This requires the
property E(tx)
= (E(t))^{x} for any real t and x,
which in turn requires the
property x L(t) = L(t ^{x}), which we now show.
We have (d/dt)
(x
L(t)
– L(t ^{x})) = x/t – (1/t ^{x})(xt ^{x}^{–1}) = x /t – x /t = 0
(it's the
derivative with respect to t, so
t ^{x}
is a power function). Consequently x L(t) – L(t ^{x})
= C, a constant. At t = 1 we have C
=
x L(1) – L(1^{x}) = x L(1) – L(1)
= x(0) – 0 = 0. This yields x L(t) = L(t ^{x}).
Now let y = E(tx).
Then L( y)
= L(E(tx)) = tx
= xt = xL(E(t)) = L((E(t))^{x}). Hence E(tx) =
y = (E(t)) ^{x}, where the last
equation
is obtained because L is
onetoone.
Since L(e) = 1, we have E(1)
= e. Then E(x) = E(1x) = (E(1))^{x} = e^{x}.
We've shown that E(x)
= e^{x}.
It follows that e^{x}
is the
inverse function of L(x). From now on we use e^{x} instead
of E(x).
Next let's find the derivative of e^{x}. Let y = e^{x}. It follows that x = L( y). Differentiating this equation
implicitly with respect to x
we get:
Indeed the limit is 1. Thus we can use the earlier finding that 2.70 < e < 2.72.
Definition 2.1  The Number e
There exists a number called the Euler constant and denoted e, with 2.70 < e < 2.72, such that: The area of the plane region bounded by the curve y = 1/x, the xaxis, the vertical line x = 1, and the vertical line x = e equals 1 square unit. This number e has the property: The existence of e guarantees that for any b > 0, b^{x} is a differentiable function. 
Remarks 2.1
1. The process doesn't assume at all that e exists. It proves, with the help of the IntermediateValue Theorem, that e exists.
2. The number e is called the Euler constant, named
after Leonhard Euler, an 18thcentury Swiss mathematician. The name
“Euler” is pronounced “oiler”.
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3. Representation Of e As A Limit 
Let's calculate the value of:
for some larger and larger values of the positive integer n. First remark that as the exponent n gets larger and larger, the
_{} 
Fig. 3.1 Approximate Values Of (1 + (1/n))^{n} For Some 
base 1 + 1/n gets smaller and smaller towards 1. Now using our calculator we have the table in Fig. 3.1. Apparently:
For the proof of the following theorem, we use L(x), which is the
inverse of e^{x}, and the
property x L(t) = L(t ^{x}), as discussed
earlier under the heading It’s Our Desired e.
Theorem 3.1 – e As A Limit

Proof
_{}
= 1.
Since L is differentiable (it has derivative 1/x), it's continuous. So we get:
EOP
The following theorem generalizes the above theorem to e^{x}.
Theorem 3.2 – e^{x} As A Limit
For any real number x we have: 
Proof
= x.
Since L is differentiable (it has derivative 1/x), it's continuous. So we get:
EOP
Remark 3.1
Eq. [3.1] can be obtained from Eq. [3.2] by substituting x = 1 in Eq. [3.2]. Eq. [3.1] is a special case of Eq. [3.2].
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4. Approximations Of e 
Eq. 3.1, reproduced here for convenience:
can be used to compute approximations of e, as shown in the table in Fig. 4.1,
which, for convenience, is a reproduction of
_{} 
Fig 4.1 Approximate Values Of (1 + (1/n))^{n} For Some 
the table in Fig 3.1. For example,
the value 2.718 28, obtained when n = 1
000 000, can be used as an approximate value of e
accurate to 5 decimal places. Of course the larger n
is, the better the approximation is.
We'll see in Section 15.3 Part 3 that:
Just the first 10 terms of the series already give a more accurate
approximation of e than the 1 000 000th
term of the
sequence (1 + 1/n)^{n}
does!! It can be proved that the series yields a better approximation of e than the sequence (1 + 1/n)^{n}
does.
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5. The Natural Exponential Function 
The function e^{x} is a particular exponential
function: it's the exponential function of base e.
The number e is very important in
mathematics and its applications. For this reason, e^{x} is called the natural exponential function.
Definition 5.1 – The Natural Exponential Function
The exponential function y = e^{x} of base e is called the natural exponential function. For the sake of simplicity, it's often called just the exponential function. In contexts where there's no ambiguity, when an exponential function is referred to without specifying the base, it's understood to be the natural exponential function. 
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6. Laws Of Exponents 
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.
The proof of the following theorem requires 2 additional
properties of L(x),
which are L(st)
= L(s)
+ L(t)
and L(s/t) = L(s) 
L(t),
which we now prove.
For L(st) = L(s) + L(t). We have (d/dt) (L(st)  L(t)) = (d/dt) L(st)  (d/dt) L(t) = (1/st)s – 1/t =
0. So L(st)
 L(t)
= C for
all t > 0, where C is some constant. In particular, for t = 1 we get L(s . 1)  L(1)
= C, or L(s)  0 = C,
yielding C = L(s).
Thus L(st)
 L(t)
= L(s),
or L(st)
= L(s)
+ L(t).
Recall from under the heading It’s Our Desired e the property x L(t) = L(t ^{x}). Consequently L(1/t) = L(t ^{1}) = 1L(t) = L(t).
For L(s/t) = L(s)  L(t). We have L(s/t) = L(s(1/t)) = L(s) + L(1/t) = L(s)  L(t).
Theorem 6.1  Laws Of Exponents
a. e^{x}e^{y} = e^{x}^{+}^{y}. e. e ^{0} = 1. 
Proof
a. L(e^{x}e^{y}) = L(e^{x})
+ L(e^{y}) = x + y = L(e^{x}^{+}^{y}).
So e^{x}e^{y} = e^{x}^{+}^{y},
since L
is invertible and thus onetoone.
b. L(e ^{–}^{x})
= x = L(e^{x})
= L(1/e^{x}). Consequently
e^{–}^{x} = 1/e^{x}.
c. L(e^{x}/e^{y}) = L(e^{x})
 L(e^{y}) = x  y = L(e^{x}^{}^{y}). Hence e^{x}/e^{y} = e^{x}^{}^{y}.
d. L(e^{x}) ^{y} = yL(e^{x}) = yx = xy = L(e^{xy}).
It follows that (e^{x})^{y}
= e^{xy}.
e. e ^{0} = e^{1–1} = e^{1}/e^{1} = 1.
EOP
Example 6.1
Simplify the expression:
EOS
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7. Differentiation 
For formality and for now that e has been shown to exist, we restate as a theorem with proof the fact that (d/dx) e^{x} = e^{x}.
Theorem 7.1 – Derivative Of e^{x}^{}
The derivative of the natural exponential function e^{x} is e^{x} itself: 
Proof
Let y = e^{x}. So x = L( y). Differentiating this equation
implicitly with respect to x we
get:
EOP
Remarks 7.1
1. Again note
that (d/dx)
e^{x}
is not xe^{x}^{–1}!! We have (d/dx) x^{t} = tx^{t}^{–1}, while (d/dx) e^{x}
= e^{x}.
The power function x^{t} has the base
as the variable and the exponent as
the constant, while the exponential function e^{x} has the base as the
constant and the
exponent as the variable.
2. Let y = e^{x}. Then y '
= e^{x},
y '' = e^{x}, y ''' = e^{x}, y ^{(4)} = e^{x}, y ^{(5)}
= e^{x},
etc. Since the (first) derivative of e^{x} is e^{x}
itself, the derivative
of any order of e^{x} is e^{x}
itself.
Example 7.1
Differentiate y = e^{sin}^{ }^{3}^{x}.
Solution
dy/dx = e^{sin}^{ }^{3}^{x} (d/dx) sin 3x = e^{sin}^{ }^{3}^{x} (cos 3x)(d/dx) 3x = e^{sin}^{ }^{3}^{x} (cos 3x)(3) = 3e^{sin}^{ }^{3}^{x} cos 3x.
EOS
Example 7.2
Find the derivative of any order of e^{ax}^{+}^{b} 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 by using
mathematical induction.
Solution
Thus the formula is also true for n + 1. Consequently, by mathematical induction, it's true for all positive integers.
EOS
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8. Graph 
Inflection Points. Since there's no change of concavity, there's no inflection point.
We use a calculator to calculate the values of y = e^{x} for some values of x, as displayed in Fig. 8.1. To find e in a calculator,
press the button e, or calculate e^{1},
or calculate (inverse of ln)(1), then store it in memory for use. We plot the
points from
Fig. 8.1 Table Of Values For y = e^{x}. 
Fig. 8.2 Graphs Of y =
e^{x}
And Its Tangent y = x + 1 At yIntercept
(0, 1). The Slope 
Fig. 8.1 in Fig. 8.2, and join them, according to the properties
discussed above, by a smooth curve to obtain the graph of y
=
e^{x}.
The slope of the tangent to the graph of y = e^{x} at the yintercept (0, 1) is y ’(0) = e^{x}_{x}_{=0} = e^{0} = 1. So the equation of this
tangent is y = x
+ 1. The tangent is also sketched in Fig. 8.2.
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9. Fastness 
Refer to Fig. 9.1. When x is negative
and increases from near minus infinity to 0, e^{x} increases only from
the near above of 0 to
1. But when x is positive and increases, e^{x}
increases very fast. In fact, it increases faster and faster, as evidenced by
the fact
that its graph becomes steeper and steeper.
The limit:
Fig. 9.1 Graphs Of y = e^{x}. 
that this limit is infinity, meaning that e^{x} approaches being infinitely larger than x^{n} as x approaches infinity.
Let's compare the sizes of e^{x}
and x^{1,000}. Surely at x = 2 we have e^{2} < 2^{1,000}, and at x = 3 we still have e^{3} < 3^{1,000}. Now
consider
x = 10,000. Of course e^{10,000}
has 10,000 factors e’s
while 10,000^{1,000} has only
1,000 factors 10,000’s. So, although e
< 10,000
by a large margin, it’s possible that e^{10,000} > 10,000^{1,000}. With the help of our calculator we
get:
The proof of the following theorem makes use of this property of L(x):
Next we show that L(x) < x^{t}/t
for all x > 0 and for any t > 0. We have, by the above
paragraph, L(x^{t}) < x^{t}.
Therefore, by the
discussion under the heading It’s Our Desired e, tL(x)
< x^{t}.
So, since t > 0, we get L(x)
< x^{t}/t.
Now we prove the above limit. By the above paragraph we have:
Theorem 9.1  Fastness Of Growth Of e^{x}^{}
For any positive integer n: So, given any positiveinteger power of x, when x
is positive and gets larger and larger, from some point on e^{x}
grows faster 
Proof
EOP
Remarks 9.1
a. Again the
limit [10.1] means that although both e^{x} and x^{n} approach infinity as x does, e^{x} approaches being infinitely larger
than x^{n}.
b. For a given integer n
> 0, no matter how large it is, e^{x}
may be smaller than x^{n}
only for x in a bounded 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 x^{n}
does.
c. This “ fastnessofgrowth” property of e^{x}
has led to the popular “exponential” phrases such as “exponential growth” or
“increase exponentially”.
d. Eq. [9.1] can be written as:
Problems & Solutions 
1. Simplify
the expression:
Solution
2. 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 by using
mathematical induction.
Solution
a. y' = e^{ax} + axe^{ax},
y'' = ae^{ax} + ae^{ax} + a^{2}xe^{ax} = 2ae^{ax} + a^{2}xe^{ax},
y''' = 2a^{2}e^{ax} + a^{2}e^{ax} + a^{3}xe^{ax} = 3a^{2}e^{ax} + a^{3}xe^{ax}.
b. y ^{(}^{n}^{)} = na^{n}^{–1}e^{ax} + a^{n}xe^{ax}.
c. As y' = e^{ax} + axe^{ax}, the formula
is true for n
= 1. Assume it's true for some positive integer n, so that y ^{(}^{n}^{)} = na^{n}^{–1}e^{ax} +
a^{n}xe^{ax}. Then:
y ^{(}^{n}^{+1)} = (d/dx) y ^{(}^{n}^{)}
= na^{n}e^{ax} + a^{n}e^{ax} + a^{n}^{+1}xe^{ax}
= (n +
1)a^{n}e^{ax} + a^{n}^{+1}xe^{ax}
= (n +
1)a^{(}^{n}^{+1)–1}e^{ax} + a^{n}^{+1}xe^{ax}.
Thus the formula is also true for n + 1. Consequently, by mathematical induction, it's true for all positive integers.
3. 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.
4. The function L(x)
defined in this section is the inverse of the natural exponential function e^{x}
and is called the natural
logarithm and denoted ln x,
where “ln”
is pronounced like the word “lawn”. So ln x
is an antiderivative of 1/x, ln e = 1,
and ln x^{t} = t ln x
for all x > 0 and for any t. Find the slope of the curve:
Solution
Differentiating the given equation implicitly with respect to x we get:
5. Prove that y ' = ky iff y
= Ce^{kx}, where y = f (x) is a function and k
and C are constants.
Hint: To show that y '
= ky implies y = Ce^{kx},
you can show that y/e^{kx} is a constant function.
Note: So a changing quantity has its
rate of change proportional to the quantity itself iff the quantity is an
exponential
function.
Solution
Suppose y ' = ky. Then:
So y/e^{kx} = C, a constant. Thus y = Ce^{kx}.
Conversely, suppose y = Ce^{kx}. Then y ' = Cke^{kx} = k(Ce^{kx}) = ky.
Consequently, y ' = ky iff y = Ce^{kx}.
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