Transcendental functions such as exponential, trigonometric and hyperbolic functions (Part - 2) | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


4.6 Exponential and Logarithmi
 fun
tions An exponential function has the form a
x
, where a is a constant; examples are 2
x
, 10
x
, e
x
.
The logarithmic functions are the inverses of the exponential functions, that is, functions
that “undo” the exponential functions, just as, for example, the cube root function “un-
does” the cube function:
3
v
2
3
= 2. Note that the original function also undoes the inverse
function: (
3
v
8)
3
= 8.
Let f(x) = 2
x
. The inverse of this function is called the logarithm base 2, denoted
log
2
(x) or (especially in computer science circles) lg(x). What does this really mean? The
logarithmmust undo the actionof theexponential function, sofor exampleit must bethat
lg(2
3
) = 3—starting with 3, the exponential function produces 2
3
= 8, and the logarithm
of 8 must get us back to 3. A little thought shows that it is not a coincidence that lg(2
3
)
simply gives the exponent—the exponent is the original value that we must get back to.
In other words, the logarithm is the exponent. Remember this catchphrase, and what it
means, and you won’t go wrong. (You do have to remember what it means. Like any
good mnemonic, “the logarithm is the exponent” leaves out a lot of detail, like “Which
exponent?” and “Exponent of what?”)
EXAMPLE 4.6.1 What is the value of log
10
(1000)? The “10” tells us the appropriate
number to use for the base of the exponential function. The logarithm is the exponent,
so the question is, what exponent E makes 10
E
= 1000? If we can ?nd such an E, then
log
10
(1000) = log
10
(10
E
) = E; ?nding the appropriate exponent is the same as ?nding the
logarithm. In this case, of course, it is easy: E = 3 so log
10
(1000)= 3.
Let’s review some laws of exponents and logarithms; let a be a positive number. Since
a
5
= a·a·a·a·aand a
3
= a·a·a,it’s clear that a
5
·a
3
= a·a·a·a·a·a·a·a= a
8
= a
5+3
,
and in general that a
m
a
n
= a
m+n
. Since “the logarithm is the exponent,” it’s no surprise
that this translates directly into a fact about the logarithm function. Here are three facts
Page 2


4.6 Exponential and Logarithmi
 fun
tions An exponential function has the form a
x
, where a is a constant; examples are 2
x
, 10
x
, e
x
.
The logarithmic functions are the inverses of the exponential functions, that is, functions
that “undo” the exponential functions, just as, for example, the cube root function “un-
does” the cube function:
3
v
2
3
= 2. Note that the original function also undoes the inverse
function: (
3
v
8)
3
= 8.
Let f(x) = 2
x
. The inverse of this function is called the logarithm base 2, denoted
log
2
(x) or (especially in computer science circles) lg(x). What does this really mean? The
logarithmmust undo the actionof theexponential function, sofor exampleit must bethat
lg(2
3
) = 3—starting with 3, the exponential function produces 2
3
= 8, and the logarithm
of 8 must get us back to 3. A little thought shows that it is not a coincidence that lg(2
3
)
simply gives the exponent—the exponent is the original value that we must get back to.
In other words, the logarithm is the exponent. Remember this catchphrase, and what it
means, and you won’t go wrong. (You do have to remember what it means. Like any
good mnemonic, “the logarithm is the exponent” leaves out a lot of detail, like “Which
exponent?” and “Exponent of what?”)
EXAMPLE 4.6.1 What is the value of log
10
(1000)? The “10” tells us the appropriate
number to use for the base of the exponential function. The logarithm is the exponent,
so the question is, what exponent E makes 10
E
= 1000? If we can ?nd such an E, then
log
10
(1000) = log
10
(10
E
) = E; ?nding the appropriate exponent is the same as ?nding the
logarithm. In this case, of course, it is easy: E = 3 so log
10
(1000)= 3.
Let’s review some laws of exponents and logarithms; let a be a positive number. Since
a
5
= a·a·a·a·aand a
3
= a·a·a,it’s clear that a
5
·a
3
= a·a·a·a·a·a·a·a= a
8
= a
5+3
,
and in general that a
m
a
n
= a
m+n
. Since “the logarithm is the exponent,” it’s no surprise
that this translates directly into a fact about the logarithm function. Here are three facts
from the example: log
a
(a
5
) = 5, log
a
(a
3
) = 3, log
a
(a
8
) = 8. So log
a
(a
5
a
3
) = log
a
(a
8
) =
8 = 5+3 = log
a
(a
5
)+log
a
(a
3
). Now let’s make this a bit more general. Suppose A and
B are two numbers, A = a
x
, and B = a
y
. Then log
a
(AB) = log
a
(a
x
a
y
) = log
a
(a
x+y
) =
x+y = log
a
(A)+log
a
(B).
Now consider (a
5
)
3
= a
5
·a
5
·a
5
= a
5+5+5
= a
5·3
= a
15
. Again it’s clear that more
generally (a
m
)
n
= a
mn
, and again this gives us a fact about logarithms. If A = a
x
then
A
y
= (a
x
)
y
= a
xy
, so log
a
(A
y
) = xy = ylog
a
(A)—the exponent can be “pulled out in
front.”
Wehavecheatedabitintheprevioustwoparagraphs. Itisobviousthata
5
=a·a·a·a·a
and a
3
=a·a·a and that the rest of the example follows; likewise for the second example.
But when we consider an exponential function a
x
we can’t be limited to substituting
integers for x. What does a
2.5
or a
-1.3
or a
p
mean? And is it really true that a
2.5
a
-1.3
=
a
2.5-1.3
? The answer to the ?rst question is actually quite di?cult, so we will evade it;
the answer to the second question is “yes.”
We’llevadethefullanswertothehardquestion, butwehavetoknowsomethingabout
exponential functions. You need ?rst to understand that since it’s not “obvious” what 2
x
should mean, we are really free to make it mean whatever we want, so long as we keep the
behavior that is obvious, namely, when x is a positive integer. What else do we want to
be true about 2
x
? We want the properties of the previous two paragraphs to be true for
all exponents: 2
x
2
y
= 2
x+y
and (2
x
)
y
= 2
xy
.
After the positive integers, the next easiest number to understand is 0: 2
0
= 1. You
have presumably learned this fact in the past; why is it true? It is true precisely because
we want 2
a
2
b
= 2
a+b
to be true about the function 2
x
. We need it to be true that
2
0
2
x
= 2
0+x
= 2
x
, and this only works if 2
0
= 1. The same argument implies that a
0
= 1
for any a.
The next easiest set of numbers to understand is the negative integers: for example,
2
-3
= 1/2
3
. We know that whatever 2
-3
means it must be that 2
-3
2
3
= 2
-3+3
= 2
0
= 1,
which means that 2
-3
must be 1/2
3
. In fact, by the same argument, once we know what
2
x
means for some value of x, 2
-x
must be 1/2
x
and more generally a
-x
= 1/a
x
.
Next, consider an exponent 1/q, where q is a positive integer. We want it to be true
that (2
x
)
y
= 2
xy
, so(2
1/q
)
q
= 2. Thismeans that 2
1/q
isaq-throot of2, 2
1/q
=
q
v
2 . This
is all we need to understand that 2
p/q
= (2
1/q
)
p
= (
q
v
2 )
p
and a
p/q
= (a
1/q
)
p
= (
q
v
a )
p
.
What’s left is the hard part: what does 2
x
mean when x cannot be written as a
fraction, like x =
v
2 or x = p? What we know so far is how to assign meaning to 2
x
Page 3


4.6 Exponential and Logarithmi
 fun
tions An exponential function has the form a
x
, where a is a constant; examples are 2
x
, 10
x
, e
x
.
The logarithmic functions are the inverses of the exponential functions, that is, functions
that “undo” the exponential functions, just as, for example, the cube root function “un-
does” the cube function:
3
v
2
3
= 2. Note that the original function also undoes the inverse
function: (
3
v
8)
3
= 8.
Let f(x) = 2
x
. The inverse of this function is called the logarithm base 2, denoted
log
2
(x) or (especially in computer science circles) lg(x). What does this really mean? The
logarithmmust undo the actionof theexponential function, sofor exampleit must bethat
lg(2
3
) = 3—starting with 3, the exponential function produces 2
3
= 8, and the logarithm
of 8 must get us back to 3. A little thought shows that it is not a coincidence that lg(2
3
)
simply gives the exponent—the exponent is the original value that we must get back to.
In other words, the logarithm is the exponent. Remember this catchphrase, and what it
means, and you won’t go wrong. (You do have to remember what it means. Like any
good mnemonic, “the logarithm is the exponent” leaves out a lot of detail, like “Which
exponent?” and “Exponent of what?”)
EXAMPLE 4.6.1 What is the value of log
10
(1000)? The “10” tells us the appropriate
number to use for the base of the exponential function. The logarithm is the exponent,
so the question is, what exponent E makes 10
E
= 1000? If we can ?nd such an E, then
log
10
(1000) = log
10
(10
E
) = E; ?nding the appropriate exponent is the same as ?nding the
logarithm. In this case, of course, it is easy: E = 3 so log
10
(1000)= 3.
Let’s review some laws of exponents and logarithms; let a be a positive number. Since
a
5
= a·a·a·a·aand a
3
= a·a·a,it’s clear that a
5
·a
3
= a·a·a·a·a·a·a·a= a
8
= a
5+3
,
and in general that a
m
a
n
= a
m+n
. Since “the logarithm is the exponent,” it’s no surprise
that this translates directly into a fact about the logarithm function. Here are three facts
from the example: log
a
(a
5
) = 5, log
a
(a
3
) = 3, log
a
(a
8
) = 8. So log
a
(a
5
a
3
) = log
a
(a
8
) =
8 = 5+3 = log
a
(a
5
)+log
a
(a
3
). Now let’s make this a bit more general. Suppose A and
B are two numbers, A = a
x
, and B = a
y
. Then log
a
(AB) = log
a
(a
x
a
y
) = log
a
(a
x+y
) =
x+y = log
a
(A)+log
a
(B).
Now consider (a
5
)
3
= a
5
·a
5
·a
5
= a
5+5+5
= a
5·3
= a
15
. Again it’s clear that more
generally (a
m
)
n
= a
mn
, and again this gives us a fact about logarithms. If A = a
x
then
A
y
= (a
x
)
y
= a
xy
, so log
a
(A
y
) = xy = ylog
a
(A)—the exponent can be “pulled out in
front.”
Wehavecheatedabitintheprevioustwoparagraphs. Itisobviousthata
5
=a·a·a·a·a
and a
3
=a·a·a and that the rest of the example follows; likewise for the second example.
But when we consider an exponential function a
x
we can’t be limited to substituting
integers for x. What does a
2.5
or a
-1.3
or a
p
mean? And is it really true that a
2.5
a
-1.3
=
a
2.5-1.3
? The answer to the ?rst question is actually quite di?cult, so we will evade it;
the answer to the second question is “yes.”
We’llevadethefullanswertothehardquestion, butwehavetoknowsomethingabout
exponential functions. You need ?rst to understand that since it’s not “obvious” what 2
x
should mean, we are really free to make it mean whatever we want, so long as we keep the
behavior that is obvious, namely, when x is a positive integer. What else do we want to
be true about 2
x
? We want the properties of the previous two paragraphs to be true for
all exponents: 2
x
2
y
= 2
x+y
and (2
x
)
y
= 2
xy
.
After the positive integers, the next easiest number to understand is 0: 2
0
= 1. You
have presumably learned this fact in the past; why is it true? It is true precisely because
we want 2
a
2
b
= 2
a+b
to be true about the function 2
x
. We need it to be true that
2
0
2
x
= 2
0+x
= 2
x
, and this only works if 2
0
= 1. The same argument implies that a
0
= 1
for any a.
The next easiest set of numbers to understand is the negative integers: for example,
2
-3
= 1/2
3
. We know that whatever 2
-3
means it must be that 2
-3
2
3
= 2
-3+3
= 2
0
= 1,
which means that 2
-3
must be 1/2
3
. In fact, by the same argument, once we know what
2
x
means for some value of x, 2
-x
must be 1/2
x
and more generally a
-x
= 1/a
x
.
Next, consider an exponent 1/q, where q is a positive integer. We want it to be true
that (2
x
)
y
= 2
xy
, so(2
1/q
)
q
= 2. Thismeans that 2
1/q
isaq-throot of2, 2
1/q
=
q
v
2 . This
is all we need to understand that 2
p/q
= (2
1/q
)
p
= (
q
v
2 )
p
and a
p/q
= (a
1/q
)
p
= (
q
v
a )
p
.
What’s left is the hard part: what does 2
x
mean when x cannot be written as a
fraction, like x =
v
2 or x = p? What we know so far is how to assign meaning to 2
x
whenever x = p/q; if we were to graph this we’d see something like this:
.............................................................................................................................................................................................................................................................................................................................................................................................................................. .......... ..... .... .... ... ... .. ... .. .. .. .. .. . .. .. .. . .. . .. . .. . . .. . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
But this is a poor picture, because you can’t see that the “curve” is really a whole lot
of individual points, above the rational numbers on the x-axis. There are really a lot of
“holes” in the curve, above x = p, for example. But (this is the hard part) it is possible
to prove that the holes can be “?lled in”, and that the resulting function, called 2
x
, really
does have the properties we want, namely that 2
x
2
y
= 2
x+y
and (2
x
)
y
= 2
xy
.
.
4.7 Deriv a tives of the exponential and logarithmi
 fun
tions As with the sine, we don’t know anything about derivatives that allows us to compute
the derivatives of the exponential and logarithmic functions without going back to basics.
Let’s do a little work with the de?nition again:
d
dx
a
x
= lim
?x?0
a
x+?x
-a
x
?x
= lim
?x?0
a
x
a
?x
-a
x
?x
= lim
?x?0
a
x
a
?x
-1
?x
= a
x
lim
?x?0
a
?x
-1
?x
Page 4


4.6 Exponential and Logarithmi
 fun
tions An exponential function has the form a
x
, where a is a constant; examples are 2
x
, 10
x
, e
x
.
The logarithmic functions are the inverses of the exponential functions, that is, functions
that “undo” the exponential functions, just as, for example, the cube root function “un-
does” the cube function:
3
v
2
3
= 2. Note that the original function also undoes the inverse
function: (
3
v
8)
3
= 8.
Let f(x) = 2
x
. The inverse of this function is called the logarithm base 2, denoted
log
2
(x) or (especially in computer science circles) lg(x). What does this really mean? The
logarithmmust undo the actionof theexponential function, sofor exampleit must bethat
lg(2
3
) = 3—starting with 3, the exponential function produces 2
3
= 8, and the logarithm
of 8 must get us back to 3. A little thought shows that it is not a coincidence that lg(2
3
)
simply gives the exponent—the exponent is the original value that we must get back to.
In other words, the logarithm is the exponent. Remember this catchphrase, and what it
means, and you won’t go wrong. (You do have to remember what it means. Like any
good mnemonic, “the logarithm is the exponent” leaves out a lot of detail, like “Which
exponent?” and “Exponent of what?”)
EXAMPLE 4.6.1 What is the value of log
10
(1000)? The “10” tells us the appropriate
number to use for the base of the exponential function. The logarithm is the exponent,
so the question is, what exponent E makes 10
E
= 1000? If we can ?nd such an E, then
log
10
(1000) = log
10
(10
E
) = E; ?nding the appropriate exponent is the same as ?nding the
logarithm. In this case, of course, it is easy: E = 3 so log
10
(1000)= 3.
Let’s review some laws of exponents and logarithms; let a be a positive number. Since
a
5
= a·a·a·a·aand a
3
= a·a·a,it’s clear that a
5
·a
3
= a·a·a·a·a·a·a·a= a
8
= a
5+3
,
and in general that a
m
a
n
= a
m+n
. Since “the logarithm is the exponent,” it’s no surprise
that this translates directly into a fact about the logarithm function. Here are three facts
from the example: log
a
(a
5
) = 5, log
a
(a
3
) = 3, log
a
(a
8
) = 8. So log
a
(a
5
a
3
) = log
a
(a
8
) =
8 = 5+3 = log
a
(a
5
)+log
a
(a
3
). Now let’s make this a bit more general. Suppose A and
B are two numbers, A = a
x
, and B = a
y
. Then log
a
(AB) = log
a
(a
x
a
y
) = log
a
(a
x+y
) =
x+y = log
a
(A)+log
a
(B).
Now consider (a
5
)
3
= a
5
·a
5
·a
5
= a
5+5+5
= a
5·3
= a
15
. Again it’s clear that more
generally (a
m
)
n
= a
mn
, and again this gives us a fact about logarithms. If A = a
x
then
A
y
= (a
x
)
y
= a
xy
, so log
a
(A
y
) = xy = ylog
a
(A)—the exponent can be “pulled out in
front.”
Wehavecheatedabitintheprevioustwoparagraphs. Itisobviousthata
5
=a·a·a·a·a
and a
3
=a·a·a and that the rest of the example follows; likewise for the second example.
But when we consider an exponential function a
x
we can’t be limited to substituting
integers for x. What does a
2.5
or a
-1.3
or a
p
mean? And is it really true that a
2.5
a
-1.3
=
a
2.5-1.3
? The answer to the ?rst question is actually quite di?cult, so we will evade it;
the answer to the second question is “yes.”
We’llevadethefullanswertothehardquestion, butwehavetoknowsomethingabout
exponential functions. You need ?rst to understand that since it’s not “obvious” what 2
x
should mean, we are really free to make it mean whatever we want, so long as we keep the
behavior that is obvious, namely, when x is a positive integer. What else do we want to
be true about 2
x
? We want the properties of the previous two paragraphs to be true for
all exponents: 2
x
2
y
= 2
x+y
and (2
x
)
y
= 2
xy
.
After the positive integers, the next easiest number to understand is 0: 2
0
= 1. You
have presumably learned this fact in the past; why is it true? It is true precisely because
we want 2
a
2
b
= 2
a+b
to be true about the function 2
x
. We need it to be true that
2
0
2
x
= 2
0+x
= 2
x
, and this only works if 2
0
= 1. The same argument implies that a
0
= 1
for any a.
The next easiest set of numbers to understand is the negative integers: for example,
2
-3
= 1/2
3
. We know that whatever 2
-3
means it must be that 2
-3
2
3
= 2
-3+3
= 2
0
= 1,
which means that 2
-3
must be 1/2
3
. In fact, by the same argument, once we know what
2
x
means for some value of x, 2
-x
must be 1/2
x
and more generally a
-x
= 1/a
x
.
Next, consider an exponent 1/q, where q is a positive integer. We want it to be true
that (2
x
)
y
= 2
xy
, so(2
1/q
)
q
= 2. Thismeans that 2
1/q
isaq-throot of2, 2
1/q
=
q
v
2 . This
is all we need to understand that 2
p/q
= (2
1/q
)
p
= (
q
v
2 )
p
and a
p/q
= (a
1/q
)
p
= (
q
v
a )
p
.
What’s left is the hard part: what does 2
x
mean when x cannot be written as a
fraction, like x =
v
2 or x = p? What we know so far is how to assign meaning to 2
x
whenever x = p/q; if we were to graph this we’d see something like this:
.............................................................................................................................................................................................................................................................................................................................................................................................................................. .......... ..... .... .... ... ... .. ... .. .. .. .. .. . .. .. .. . .. . .. . .. . . .. . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
But this is a poor picture, because you can’t see that the “curve” is really a whole lot
of individual points, above the rational numbers on the x-axis. There are really a lot of
“holes” in the curve, above x = p, for example. But (this is the hard part) it is possible
to prove that the holes can be “?lled in”, and that the resulting function, called 2
x
, really
does have the properties we want, namely that 2
x
2
y
= 2
x+y
and (2
x
)
y
= 2
xy
.
.
4.7 Deriv a tives of the exponential and logarithmi
 fun
tions As with the sine, we don’t know anything about derivatives that allows us to compute
the derivatives of the exponential and logarithmic functions without going back to basics.
Let’s do a little work with the de?nition again:
d
dx
a
x
= lim
?x?0
a
x+?x
-a
x
?x
= lim
?x?0
a
x
a
?x
-a
x
?x
= lim
?x?0
a
x
a
?x
-1
?x
= a
x
lim
?x?0
a
?x
-1
?x
There aretwointeresting thingstonote here: As inthecase of thesine function we areleft
with a limit that involves ?x but not x, which means that whatever lim
?x?0
(a
?x
-1)/?x
is, we know that it is a number, that is, a constant. This means that a
x
has a remarkable
property: its derivative is a constant times itself.
We earlier remarked that the hardest limit we would compute is lim
x?0
sinx/x = 1; we
now have a limit that is just a bit too hard to include here. In fact the hard part is to
see that lim
?x?0
(a
?x
-1)/?x even exists—does this fraction really get closer and closer to
some ?xed value? Yes it does, but we will not prove this fact.
We can look at some examples. Consider (2
x
-1)/x for some small values of x: 1,
0.828427124, 0.756828460, 0.724061864, 0.70838051, 0.70070877 when x is 1, 1/2, 1/4,
1/8, 1/16, 1/32, respectively. It looks like this is settling in around 0.7, which turns out
to be true (but the limit is not exactly 0.7). Consider next (3
x
-1)/x: 2, 1.464101616,
1.264296052, 1.177621520, 1.13720773, 1.11768854, at the same values of x. It turns out
to be true that in the limit this is about 1.1. Two examples don’t establish a pattern, but
if you do more examples you will ?nd that the limit varies directly with the value of a:
bigger a, bigger limit; smaller a, smaller limit. As we can already see, some of these limits
will be less than 1 and some larger than 1. Somewhere between a = 2 and a = 3 the limit
will be exactly 1; the value at which this happens is called e, so that
lim
?x?0
e
?x
-1
?x
= 1.
Asyou might guess fromour twoexamples, e iscloser to3than to2, and infact e˜ 2.718.
Now we see that the function e
x
has a truly remarkable property:
d
dx
e
x
= lim
?x?0
e
x+?x
-e
x
?x
= lim
?x?0
e
x
e
?x
-e
x
?x
= lim
?x?0
e
x
e
?x
-1
?x
= e
x
lim
?x?0
e
?x
-1
?x
= e
x
That is, e
x
is its own derivative, or in other words the slope of e
x
is the same as its height,
or the same as its second coordinate: The function f(x) = e
x
goes through the point
(z,e
z
) and has slope e
z
there, no matter what z is. It is sometimes convenient to express
the function e
x
without an exponent, since complicated exponents can be hard to read. In
such cases we use exp(x), e.g., exp(1+x
2
) instead of e
1+x
2
.
Page 5


4.6 Exponential and Logarithmi
 fun
tions An exponential function has the form a
x
, where a is a constant; examples are 2
x
, 10
x
, e
x
.
The logarithmic functions are the inverses of the exponential functions, that is, functions
that “undo” the exponential functions, just as, for example, the cube root function “un-
does” the cube function:
3
v
2
3
= 2. Note that the original function also undoes the inverse
function: (
3
v
8)
3
= 8.
Let f(x) = 2
x
. The inverse of this function is called the logarithm base 2, denoted
log
2
(x) or (especially in computer science circles) lg(x). What does this really mean? The
logarithmmust undo the actionof theexponential function, sofor exampleit must bethat
lg(2
3
) = 3—starting with 3, the exponential function produces 2
3
= 8, and the logarithm
of 8 must get us back to 3. A little thought shows that it is not a coincidence that lg(2
3
)
simply gives the exponent—the exponent is the original value that we must get back to.
In other words, the logarithm is the exponent. Remember this catchphrase, and what it
means, and you won’t go wrong. (You do have to remember what it means. Like any
good mnemonic, “the logarithm is the exponent” leaves out a lot of detail, like “Which
exponent?” and “Exponent of what?”)
EXAMPLE 4.6.1 What is the value of log
10
(1000)? The “10” tells us the appropriate
number to use for the base of the exponential function. The logarithm is the exponent,
so the question is, what exponent E makes 10
E
= 1000? If we can ?nd such an E, then
log
10
(1000) = log
10
(10
E
) = E; ?nding the appropriate exponent is the same as ?nding the
logarithm. In this case, of course, it is easy: E = 3 so log
10
(1000)= 3.
Let’s review some laws of exponents and logarithms; let a be a positive number. Since
a
5
= a·a·a·a·aand a
3
= a·a·a,it’s clear that a
5
·a
3
= a·a·a·a·a·a·a·a= a
8
= a
5+3
,
and in general that a
m
a
n
= a
m+n
. Since “the logarithm is the exponent,” it’s no surprise
that this translates directly into a fact about the logarithm function. Here are three facts
from the example: log
a
(a
5
) = 5, log
a
(a
3
) = 3, log
a
(a
8
) = 8. So log
a
(a
5
a
3
) = log
a
(a
8
) =
8 = 5+3 = log
a
(a
5
)+log
a
(a
3
). Now let’s make this a bit more general. Suppose A and
B are two numbers, A = a
x
, and B = a
y
. Then log
a
(AB) = log
a
(a
x
a
y
) = log
a
(a
x+y
) =
x+y = log
a
(A)+log
a
(B).
Now consider (a
5
)
3
= a
5
·a
5
·a
5
= a
5+5+5
= a
5·3
= a
15
. Again it’s clear that more
generally (a
m
)
n
= a
mn
, and again this gives us a fact about logarithms. If A = a
x
then
A
y
= (a
x
)
y
= a
xy
, so log
a
(A
y
) = xy = ylog
a
(A)—the exponent can be “pulled out in
front.”
Wehavecheatedabitintheprevioustwoparagraphs. Itisobviousthata
5
=a·a·a·a·a
and a
3
=a·a·a and that the rest of the example follows; likewise for the second example.
But when we consider an exponential function a
x
we can’t be limited to substituting
integers for x. What does a
2.5
or a
-1.3
or a
p
mean? And is it really true that a
2.5
a
-1.3
=
a
2.5-1.3
? The answer to the ?rst question is actually quite di?cult, so we will evade it;
the answer to the second question is “yes.”
We’llevadethefullanswertothehardquestion, butwehavetoknowsomethingabout
exponential functions. You need ?rst to understand that since it’s not “obvious” what 2
x
should mean, we are really free to make it mean whatever we want, so long as we keep the
behavior that is obvious, namely, when x is a positive integer. What else do we want to
be true about 2
x
? We want the properties of the previous two paragraphs to be true for
all exponents: 2
x
2
y
= 2
x+y
and (2
x
)
y
= 2
xy
.
After the positive integers, the next easiest number to understand is 0: 2
0
= 1. You
have presumably learned this fact in the past; why is it true? It is true precisely because
we want 2
a
2
b
= 2
a+b
to be true about the function 2
x
. We need it to be true that
2
0
2
x
= 2
0+x
= 2
x
, and this only works if 2
0
= 1. The same argument implies that a
0
= 1
for any a.
The next easiest set of numbers to understand is the negative integers: for example,
2
-3
= 1/2
3
. We know that whatever 2
-3
means it must be that 2
-3
2
3
= 2
-3+3
= 2
0
= 1,
which means that 2
-3
must be 1/2
3
. In fact, by the same argument, once we know what
2
x
means for some value of x, 2
-x
must be 1/2
x
and more generally a
-x
= 1/a
x
.
Next, consider an exponent 1/q, where q is a positive integer. We want it to be true
that (2
x
)
y
= 2
xy
, so(2
1/q
)
q
= 2. Thismeans that 2
1/q
isaq-throot of2, 2
1/q
=
q
v
2 . This
is all we need to understand that 2
p/q
= (2
1/q
)
p
= (
q
v
2 )
p
and a
p/q
= (a
1/q
)
p
= (
q
v
a )
p
.
What’s left is the hard part: what does 2
x
mean when x cannot be written as a
fraction, like x =
v
2 or x = p? What we know so far is how to assign meaning to 2
x
whenever x = p/q; if we were to graph this we’d see something like this:
.............................................................................................................................................................................................................................................................................................................................................................................................................................. .......... ..... .... .... ... ... .. ... .. .. .. .. .. . .. .. .. . .. . .. . .. . . .. . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
But this is a poor picture, because you can’t see that the “curve” is really a whole lot
of individual points, above the rational numbers on the x-axis. There are really a lot of
“holes” in the curve, above x = p, for example. But (this is the hard part) it is possible
to prove that the holes can be “?lled in”, and that the resulting function, called 2
x
, really
does have the properties we want, namely that 2
x
2
y
= 2
x+y
and (2
x
)
y
= 2
xy
.
.
4.7 Deriv a tives of the exponential and logarithmi
 fun
tions As with the sine, we don’t know anything about derivatives that allows us to compute
the derivatives of the exponential and logarithmic functions without going back to basics.
Let’s do a little work with the de?nition again:
d
dx
a
x
= lim
?x?0
a
x+?x
-a
x
?x
= lim
?x?0
a
x
a
?x
-a
x
?x
= lim
?x?0
a
x
a
?x
-1
?x
= a
x
lim
?x?0
a
?x
-1
?x
There aretwointeresting thingstonote here: As inthecase of thesine function we areleft
with a limit that involves ?x but not x, which means that whatever lim
?x?0
(a
?x
-1)/?x
is, we know that it is a number, that is, a constant. This means that a
x
has a remarkable
property: its derivative is a constant times itself.
We earlier remarked that the hardest limit we would compute is lim
x?0
sinx/x = 1; we
now have a limit that is just a bit too hard to include here. In fact the hard part is to
see that lim
?x?0
(a
?x
-1)/?x even exists—does this fraction really get closer and closer to
some ?xed value? Yes it does, but we will not prove this fact.
We can look at some examples. Consider (2
x
-1)/x for some small values of x: 1,
0.828427124, 0.756828460, 0.724061864, 0.70838051, 0.70070877 when x is 1, 1/2, 1/4,
1/8, 1/16, 1/32, respectively. It looks like this is settling in around 0.7, which turns out
to be true (but the limit is not exactly 0.7). Consider next (3
x
-1)/x: 2, 1.464101616,
1.264296052, 1.177621520, 1.13720773, 1.11768854, at the same values of x. It turns out
to be true that in the limit this is about 1.1. Two examples don’t establish a pattern, but
if you do more examples you will ?nd that the limit varies directly with the value of a:
bigger a, bigger limit; smaller a, smaller limit. As we can already see, some of these limits
will be less than 1 and some larger than 1. Somewhere between a = 2 and a = 3 the limit
will be exactly 1; the value at which this happens is called e, so that
lim
?x?0
e
?x
-1
?x
= 1.
Asyou might guess fromour twoexamples, e iscloser to3than to2, and infact e˜ 2.718.
Now we see that the function e
x
has a truly remarkable property:
d
dx
e
x
= lim
?x?0
e
x+?x
-e
x
?x
= lim
?x?0
e
x
e
?x
-e
x
?x
= lim
?x?0
e
x
e
?x
-1
?x
= e
x
lim
?x?0
e
?x
-1
?x
= e
x
That is, e
x
is its own derivative, or in other words the slope of e
x
is the same as its height,
or the same as its second coordinate: The function f(x) = e
x
goes through the point
(z,e
z
) and has slope e
z
there, no matter what z is. It is sometimes convenient to express
the function e
x
without an exponent, since complicated exponents can be hard to read. In
such cases we use exp(x), e.g., exp(1+x
2
) instead of e
1+x
2
.
What about the logarithm function? This too is hard, but as the cosine function was
easier to do once the sine was done, so the logarithm is easier to do now that we know
the derivative of the exponential function. Let’s start with log
e
x, which as you probably
know is often abbreviated lnx and called the “natural logarithm” function.
Consider the relationship between the two functions, namely, that they are inverses,
thatone“undoes”theother. Graphicallythismeansthattheyhavethesamegraphexcept
that one is “?ipped” or “re?ected” through the line y =x, as shown in ?gure 4.7.1.
............................................................................................................................................................ ........... ..... ..... ... ... ... .. ... . . .. .. .. .. .. . .. . . . .. . .. . . . . .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .. . . .. . . . . .. . . . .. .. .. .. ... .. ... ... .... .... ...... ............................................................................................................................................................................
Figure 4.7.1 The exponential and logarithm functions.
This means that the slopes of these two functions are closely related as well: For example,
the slope of e
x
is e at x= 1; at the corresponding point on the ln(x) curve, the slope must
be 1/e, because the “rise” and the “run” have been interchanged. Since the slope of e
x
is
e at the point (1,e), the slope of ln(x) is 1/e at the point (e,1).
............................................................................................................................................................ ........... ..... ..... ... ... ... .. ... .. .. .. .. . . .. . .. .. . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.. . .. . . . . .. . .. . .. . . . . .. . .. . . . . .. . .. . .. . . . . .. . .. . . . .. . .. . . . . .. . .. . . . . .. . .. . .. . . . . .. . .. . . . . .. . .. . .. . . . . .. . .. . . . . .. . .. . .. . . . . .. . .. . . . . .. . .. . .. . . . . .. . .. . . . . .. . .. . .. . . . . .. .. . .. . .. . .. . . . . .. . .. . .. . .. . .. . .. . . . . .. . .. . .. . .. . .. . .. . . . . .. . .. . .. . .. . .. . .. . . . . .. . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .. . .. . .. .. . .. .. .. .. ... .. ... ... .... .... ...... ............................................................................................................................................................................
........................................................................................................................................................................................................................................................
Figure 4.7.2 Slope of the exponential and logarithm functions.
More generally, we know that the slope of e
x
is e
z
at the point (z,e
z
), so the slope of
ln(x) is 1/e
z
at (e
z
,z), as indicated in ?gure 4.7.2. In other words, the slope of lnx is the
reciprocal of the ?rst coordinate at any point; this means that the slope of lnx at (x,lnx)
is 1/x. The upshot is:
d
dx
lnx =
1
x
.