CHAPTER 24 Derivatives of Inverse Functions and Logarithms
[Pages:9]CHAPTER 24
Derivatives of Inverse Functions and Logarithms
??
We know that
d dx
x
e
= x. e
But what about derivatives of exponen? tia?l
function wit?h bas?es other? than
e??
In other words, what is
d dx
x
a
?
And what about d
and d
? The main goal of this chapter is
ln(x)
dx
dx loga(x)
to answer these quest?ion?s and thus expand our list of derivative rules.
Let's start with d x . Since
is the inverse of x, we know = ln(a).
a
ln(x)
e
ae
dx
We can thus convert the power x to a power of :
a
e
?
x = ln(a) x = ln(a)x
ae
e.
With
this,
we
can
get
the
derivative
of
x
a
with
the
chain
rule:
hi
h
i
h
i
d
x =d
ln(a)x
=
d
ln(a)x
= ln(a)x
=x
a
e
e
ln(a)x e ln(a) a ln(a).
dx
|d x
{z dx
}
chain rule
So
the
derivative
of
x
a
is
just
x
a
times
the
constant
ln(a).
This
is
a
new
rule.
hi
Rule 16
d
x
a
=
x
ln(a)a
dx
hi
hi
For example,
d
x
10
=
x
ln(10)10
?
2.302
?
x
10
.
Also
d
x
2
=
x?
ln(2)2
dx
h i dx
0.693
?
x
2
.
Notice how special the base e is:
d
x
e
=
x
ln(e)e
=
?x 1e
=
x
e
.
The base = is the only base for which the dedrixvative of x is 1 times x.
ae
??
a
a
Next we will get a rule for d ln(x) . Our strategy will be to use the fact
dx
that ln(x) is the inverse of ex, that is,
if
f (x) = ex, then
? 1
=
f (x) ln(x).
Our
pl?an
is?to
first
develop
a
general
rule
for
?
d
? 1
?
f (x)
dx
and
then
use
it
to
get d ln(x) . (See Chapter 4 If you need to review inverses.)
dx
278
Derivatives of Inverse Functions and Logarithms
h
i
Thus our immediate question is: What is
d
? 1
f (x)
?
To
answer
this,
think
about
the
relation
bedtxween
f
and its
inverse
f
? 1
:
??
1
? =
f f (x) x.
The two sides of this equation are equal functions, so if we di erentiate both sides the derivatives will be equal:
h d
?
?
1
?i
hi
=d
f f (x)
x
dx
dx
The right-hand side of this equation is 1. The left-hand side is the derivative of a composition, so we can apply the chain rule to it:
0? ?
1
?
h d
?
1
i =
f f (x) f (x) 1
dx
wIthnaeanrpteptbloyeificnnagudt!sheWewcehecaadinnonris'utolkleanwtoeewimtwbuyhltaidptilvididedxd?inff?g0 ?1tf(hx?)e1?(axisb)?.obvByeudetdxqi?utf'as?t1ei(oxxna)?cb.tyWlyfew0 ?shfto?ap1t(pxwe)de?:
h
i
d?
1
=
f (x)
0?
1
?
?
1
dx
f f (x) .
This is our latest rule.
Rule
17
(The
inverse
rule)
If
f
is
a
function
having
a
derivative
0
f
and
an
inverse
? 1
,
then
f
h
i
d?
1
=
f (x)
0?
1
?
?.
1
dx
f f (x)
Let'Tsofiinlldustdra? fte?1t(hxi)s?.ruWlee,
suppose f ( know that
x)
0
f
=
3
x
,
= (x)
which
has
an
inverse
f
? 1
(x)
3x2, so our new rule gives
=
p
3
x.
dx
h
i
d?
1
f (x)
dx
=
0?
1
?
1
?=
?
1
?
?=
2
f f (x)
1
3 f (x)
p1 .
32
3x
Granted, this is not all that impressive, since we can use the power rule to get the same answer:
h d?
1
i
hp i
=d 3 =
hi
d
1 3
=
1
?2
3
=
1
=
f (x)
x
x
x
p1 .
2
dx
dx
dx
3
32
3x 3
3x
279
But the inverse rule can be very useful. We'll now use it to find the
derivative
of
ln(x).
Say
= f (x)
x
e
,
so
f
? 1
(x)
=
ln(x).
Then
hi
h
i
d
=d
? 1
ln(x)
f (x)
dx
dx
because
ln(x)
=
f
? 1
(x)
=
0?
1
?
?
1
f f (x)
by inverse rule
= 0? 1 ? f ln(x)
because
? 1
=
f (x) ln(x)
=1
ln(x)
e
becase
0= f (x)
x
e
=1
because
ln(x)
e
=
x.
x
hi
Thus d ln(x) = 1 . Figure 24.1 (left) illustrates this remarkable fact.
It
shows
tdhxe
function
x f
(x)
=
ln(x)
along
with
its
derivative
f
0= (x)
1.
Notice
x
how if x is near 0, the tangent to ln(x) at x is very steep, and indeed the
derivative 1 is very large. But as x gets bigger, the tangent to ln(x) gets
x
closer to horizontal (slope 0) while the derivative 1 approaches zero.
x
= y ln(x)
=1 y
x
= || y ln x
=1 y
x
Figure
24.1.
Left:
the
graphs
of
= f (x) ln(x)
(black)
and
0= f (x)
1
(blue)
with
domain
(0, 1).
Right:
the
graphs
of
= || f (x) ln x
(black)
and
0
x
=
f (x)
1
(blue).
x
Notice
however,
that
the
domain
of
ln(x)
is
1 (0, )
but
the
domain
of
1
is
x
?1 (,
0)
[
(0,
1).
So
when
we
say
that
the
derivative
of
ln(x)
is
1,
we
really
x
mean 1 with its domain restricted to (0,1). Figure 24.1 (right) shows a
x
somewhat more complete scenario. It shows the function ln(|x|), which we
will
abbreviate
as
ln |x|.
This
function
has
domain
?1 (,
0)
[
(0,
1),
and
its
derivative is 1 with its usual domain. So our latest rule has two parts.
x
hi
Rule 18
d
=1
ln(x)
and
d
h ?? ??i ln x
=
1.
dx
x
dx
x
280
Derivatives of Inverse Functions and Logarithms
Here it is understood that in the first formula the domain of ln(x) and 1
x
is
(0, 1).
In
the
second
formula
the
domain
of
|| ln x
and
1
is
all
real
numbers
x
except 0. Do not sweat the di erence between the two versions of this rule ?
they say almost the same thing, and the second implies the first. We will
mostly use the first version in parts 3 and 4 of this book, but the second
version becomes particularly useful in Part 5.
At the begin?nin?g of?this ?chapter ?we said? our main goals were to find
formulas for d
dx
x
a
,
d
dx
ln(x)
and d
dx
loga(x) . We've done all but the last
one. For it we will use the change of base formula (Fact 5.1 in Chapter 5,
page 88) which states
= ln(x)
loga(x)
.
ln(a)
Using this, the constant multiple rule and Rule 18, we get
h
i
hi
d
= d ln(x) = 1 ? d
= 1 ?1
loga(x)
ln(x)
.
dx
dx ln(a) ln(a) dx
ln(a) x
With our prior agreement about domains, we get another two-part formula.
h
i
Rule 19
d
=
loga(x)
1
and
h d
?? ??i =
loga x
1.
dx
x ln(a)
dx
x ln(a)
h
i
h
i
h
i
Example 24.1
d
=d
+
d
log3(x) tan(x)
log3(x) tan(x) log3(x) tan(x)
dx
dx
dx
(product rule)
=
1
+
2
tan(x) log3(x) sec (x)
.
x ln(3)
hp
i
Example 24.2 Find d
+ 3+ 5 x ln(x)
.
dx This is the derivative of a function to a power, so we can use the generalized
power rule:
hp
i
h?
?i
d
+ 3+
= d + 3+
1/2
5 x ln(x)
5 x ln(x)
dx
dx
? = 1 + 3+
??
1/2 1
h d
+
3+
i
5 x ln(x)
5 x ln(x)
2
dx
? = 1 + 3+
?? ?
1/2
2+ 1
5 x ln(x) 3x
2
x
2+ 1
= p 3x x
+ 3+
.
2 5 x ln(x)
281
.
Example 24.3
h
? ?i
? ?h i
d ++
3 = ++
2d
7 x ln(x)
0 1 3 ln(x)
ln(x)
dx
dx
??
??
2
=+
21
1 3 ln(x)
=
+ 3 ln(x) 1
.
x
x
Example 24.4 (quotient rule)
hi
hi
h
i
d
? + ?d
ln(x) x ln(x) x
1? + ? x ln(x) 1
d ln(x) = dx
dx = x
2
2
dx x
x
x
=
+ 1 ln(x)
.
2
x
Example 24.5
Find the derivative of
2+ x3
+ x2
.
10
(
The composition =
2+ + x 3x 2
can
be
broken
up
as
y 10
=u y 10
= 2+
+
u x 3x 2.
The chain rule then gives d y = d y ? du
dx du dx
=
u? + +
ln(10) 10 (2x 3 0)
=
2+ + x 3x 2
+
ln(10) 10
(2x 3) .
In Example 24.5 we di
erentiated
a
function
of
form
a
g(x)
?
.?
Let's
repeat
our steps to get a chain rule generalization for the rule
d
x
a
=
ln(a)
x
a
.
dx
Example 24.6 Find the derivative of ag(x). (
The
composition
y
=
g(x)
a
can
be
broken
up
as
=u ya
=
u g(x).
The chain rule then gives d y = d y ? du
dx du dx
=
u? 0
ln(a) a g (x)
=
0 g(x)
ln(a) a g (x) .
d
? This?
g(x)
e
examples shws
d
??
g(x)
a
=
ln(a)ag(x) g0(x),
a
companion
to
the
rule
=
e
g(x)
0
g
(
x).
We
dx
will
summarize
these
and
chain
rule
generalizations
dx
of the other rules from this chapter on the bottom of the next page.
282
Derivatives of Inverse Functions and Logarithms
Example 24.7 Find the derivative of y = ln ?? sin(x)??.
(
This is a composition, and the function can be broken up as
= || y ln u
=
u sin(x)
The chain rule gives
dy
=
dy
du
=
1 cos(x)
=
1
= cos(x)
cos(x) .
dx du dx u
sin(x)
sin(x)
funEctxioanmopflefr2o4m.7lnil??lgu(sxt)r??aotreslna?
com?mon pattern, which is to di erentiate a g(x) . Let's redo the example in this setting.
Example 24.8 Find the derivative of y = ln??g(x)??.
(
This is a composition, and the function can be broken up as
= || y ln u
=
u g(x)
0
The chain rule gives
dy
=
dy
du
=
10 g (x)
=
1
0= g (x)
g (x) .
dx du dx u
g(x)
g(x)
Example 24.8 has shown that
h d
??
??i =
h 1 ?d
i
0
= g (x)
ln g(x)
g(x)
.
dx
g(x) dx
g(x)
??
This is the chain rule generalization of the rule d
? ? ?? 0
dx
|| ln x
= 1 , and it is worth
x
remember?ing.? It implies
d dx
ln
g(x)
= g (x) , and we often us it this way.
g(x)
(fRoreca??ll
ln
??
g(x) is not defined when g(x) is more all-encompassing.)
is
negative,
so
the
rule
as
stated
ln g(x)
Here is a summary of this chapter's main rules, along side their chain
rule generalizations. Remember them and internalize them.
Di erentiation rules for exponential and log functions
Rule
hi
d x=x ee
dxh i
d x=
x
a ln(a) a
dx
h d
?? ?? i = 1
ln x
dx
x
h d
?? ?? i = 1
loga x
dx
x ln(a)
Chain rule generalization
hi
d
g(x)
=
0 g(x)
e
e g (x)
dx
hi
d g(x) =
0 g(x)
a
ln(a) a g (x)
dx
h d
??
??i
=
0
g (x)
ln g(x)
dx
g(x)
h d
??
??i =
0
g (x)
loga g(x)
dx
g(x) ln(a)
We
prefer
the
base
e,
so
you
should
expect
to
that
the
formulas
for
x
a
and
loga to play less of a role. (Though in computer science, log2 is significant!)
283
Example 24.9
Find
the
derivative
of
y
=
?? 5 ln 4x
+ 3+ 6x
x + 3??.
We will do this in two di erent then we will use the formula d
w? ay??s. Fi??r?s=t ln g(x)
dx
we will use the chain rule, and
0
g (x) g(x)
from (the
previous
page.
Using the chain rule, we first break the function up as
= || y ln u
= 5+ 3+ +
u 4x 6x x 3
The chain rule gives d y = d y du
dx du dx
?
?
=1
4+ 2+
20x 18x 1
u
?
?
=
1
4+ 2+ =
5+ 3+ + 20x 18x 1
4x 6x x 3
4+ 2+ 20x 18x 1
5+ 3+ + . 4x 6x x 3
? Next, using the formula d
??
??
=
0
g (x) ,
from
the
previous
page,
the
ln g(x)
dx
g(x)
h answer comes in one step: d
??
5+
3+
+ ?? i =
4+ 2+ 20x 18x 1
.
ln 4x 6x x 3
5+ 3+ +
dx
4x 6x x 3
So using the formula is quicker. But you should soon reach the point
where the above to approaches are equally automatic. Doing the chain rule
in your head is in essence using the formula.
Example 24.10
Find
the
derivative
of
y
=
ln
???tan
p
2
x
+
3x
?
???.
This has the form chain rule or the
offoarmcoumlaposdit?ilonn??
ln ??g??(x=)??, g(x)
so
1
we can use either the straight
0
g (x)
from
the
previous
page.
dx
g(x)
Let's try the formula.
h d
???
p 2+
? ???i =
ln tan x 3x
p1
h p
?i
?d
2+
tan x 3x
dx
2+ dx
tan x 3x
p
? hp
i
= p1
? 2 2+ d
2+
sec x 3x
x 3x
2+
dx
tan x 3x
= p1
?
p
2
2+
?? 1 2+
?1?
1
2
d
h
2+
i
sec x 3x x 3x
x 3x
2+
2
dx
tan x 3x
= p1
?
p
2
2+
?? 1 2+
?? 1
2
+
sec x 3x x 3x (2x 3)
2+
2
tan x 3x
p
?
+
2 2+
(2x 3) sec x 3x
=
p
?p
2+
2+
2 tan x 3x x 3x
284
Derivatives of Inverse Functions and Logarithms
Exercises for Chapter 24
In exercises 1?20 di erentiate the given function.
p
?
1.
+1+ ln(x)
+ x3
2.
ln
2+ 1 x
x
x
3. ln(w)
w?
?
5.
3
ln sin (x)
p
7.
+ ?? + 5 ln( )
?3
??
9. cos ln(x) p
11.
?2 + ? + ln(5 p)
?9
13.
2+ ln(x 1)
+ 3x 1
??
15.
ln
x
xe
??
17.
tan?
ln(x)
+ x
19.
ln
+1 1
x
21.
Find
+? ln(3 h) ln(3) li!m
h0
h
4.
1 x2?+ ln(x)?
6. ln tan(x) ? ? ??
8.
ln
sec
3
x
? ? ??
10.
3
sec ln x
?
?
12.
ln
3
sec(x )
?
?
14.
sec
3
ln(x )
16.
x
ln (x) e
? ? ??
18.
ln sin
3
x
3
20.
x ln(x) 3+
22.
x Find
1 z?
28 li!m ?
z 3z 3
Exercise Solutions for Chapter 24
In exercises 1?20 di erentiate the given function.
p
1.
d
+1+ ln(x)
+ x3
=1? 1 +
p1
2
dx
x
x x 2x
3.
d
ln(w) =
1 w
??
?
w ln(w) 1
=
? 1 ln(w)
2
2
dw w
w
w
h?
?i
h
i
h
i
5.
d
3
ln sin (x)
=
1
d
3 =1
d
2
= cos(x)
sin (x)
3 sin (x) sin(x) 3
3
3
dx
sin (x) dx
sin (x)
dx
sin(x)
7.
h
pi
h
i
?
p
d ?
5 + ln(??) +
?3
=
d ?
5 + ln(??) + ?3/2
=
+ 0
??
+
3 ?1/2
=
1 ?
+
3
?
d
d
2
2
h ? ?i
??
??
9. d cos ln(x) = ? sin ln(x) 1 = ? sin ln(x)
dx
x
x
h
pi
h
i
p
11.
d
?2
+
? ln(5 )
+
?9
=d
?2
+
? ln(5 )
+
?9/2
=+ 0
5 + 9 ?7/2 = 1 + 9
?
?
?7
dx
dx
52
2
................
................
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