Lesson 16: Derivatives of Exponential and Logarithmic Functions
7 INVERSE FUNCTIONS. 7.4 Derivatives of Logarithmic Functions In this section, we find: The...
-
date post
20-Dec-2015 -
Category
Documents
-
view
220 -
download
0
Transcript of 7 INVERSE FUNCTIONS. 7.4 Derivatives of Logarithmic Functions In this section, we find: The...
7INVERSE FUNCTIONSINVERSE FUNCTIONS
7.4Derivatives of
Logarithmic Functions
In this section, we find:
The derivatives of the logarithmic functions y = logax
and the exponential functions y = ax.
INVERSE FUNCTIONS
We start with the natural logarithmic
function y = ln x.
We know it is differentiable because it is the inverse of the differentiable function y = ex.
DERIVATIVES OF LOG FUNCTIONS
1(ln )
dx
dx x
Formula 1
DERIVATIVES OF LOG FUNCTIONS Formula 1—Proof
Let y = ln x.
Then, ey = x.
Differentiating this equation implicitly with respect to x, we get:
So,
1y dyedx
1 1y
dy
dx e x
Differentiate y = ln(x3 + 1).
To use the Chain Rule, we let u = x3 + 1. Then, y = ln u. So,
23
2
3
1 1(3 )
1
3
1
dy dy du dux
dx du dx u dx x
x
x
DERIVATIVES OF LOG FUNCTIONS Example 1
In general, if we combine Formula 1
with the Chain Rule, as in Example 1,
we get:
1 '( )(ln ) or ln ( )
( )
d du d g xu g x
dx u dx dx g x
DERIVATIVES OF LOG FUNCTIONS Formula 2
Find .
Using Formula 2, we have:
ln(sin )d
xdx
1ln(sin ) (sin )
sin1
cos cotsin
d dx x
dx x dx
x xx
DERIVATIVES OF LOG FUNCTIONS Example 2
Differentiate .
This time, the logarithm is the inner function.
So, the Chain Rule gives:
( ) lnf x x
1 212
1 1'( ) (ln ) (ln )
2 ln1
2 ln
df x x x
dx xx
x x
DERIVATIVES OF LOG FUNCTIONS Example 3
Find .
1 212
12
1 1 1ln
12 22
2 1 ( 1) ( 2)2
1 22 ( 1) 5
( 1)( 2) 2( 1)( 2)
d x d xxdx dxx xx
x x xx
x xx x x
x x x x
1ln
2
d x
dx x
DERIVATIVES OF LOG FUNCTIONS E. g. 4—Solution 1
If we first simplify the given function using the
laws of logarithms, the differentiation becomes
easier:
This answer can be left as written. However, if we used a common denominator,
it would give the same answer as in Solution 1.
12
1ln ln( 1) ln( 2)
2
1 1 1
1 2 2
d x dx x
dx dxx
x x
DERIVATIVES OF LOG FUNCTIONS E. g. 4—Solution 2
Find the absolute minimum value of:
f(x) = x2 ln x
The domain is (0, ∞) and the Product Rule gives:
Therefore, f ’(x) = 0 when 2 ln x = -1, that is, ln x = -½, or x = e-½.
Example 5DERIVATIVES OF LOG FUNCTIONS
2 1'( ) 2 ln (1 2ln )f x x x x x x
x
Also, f ’(x) > 0 when x > e-½
and f ’(x) < 0 when 0 < x < e-½.
So, by the First Derivative Test for Absolute Extreme Values,
is the absolute minimum.
Example 5
(1/ ) 1/(2 )f e e
DERIVATIVES OF LOG FUNCTIONS
Discuss the curve
y = ln(4 – x2)
using the guidelines
of Section 4.5
DERIVATIVES OF LOG FUNCTIONS Example 6
A. The domain is:
DERIVATIVES OF LOG FUNCTIONS Example 6
2 2{ 4 0} { 4}
{ | | 2}
( 2,2)
x x x x
x x
B. The y-intercept is f(0) = ln 4.
To find the x-intercept we set:
y = ln(4 – x2) = 0
We know that ln 1 = loge1 = 0 (since e0 = 1).
So, we have 4 – x2 = 1 x2 = 3. Therefore, the x-intercepts are .
DERIVATIVES OF LOG FUNCTIONS Example 6
3
Since f(-x) = f(x), f is even
and the curve is symmetric
about the y-axis.
DERIVATIVES OF LOG FUNCTIONS Example 6
D. We look for vertical asymptotes at
the endpoints of the domain.
Since 4 – x2 → 0+ as x → 2- and also as x → -2+, we have:
by Equation 8 in Section 7.3. So, the lines x = 2 and x = -2 are vertical
asymptotes.
DERIVATIVES OF LOG FUNCTIONS Example 6
2 2
2 2lim ln(4 ) lim ln(4 )x x
x x
and
DERIVATIVES OF LOG FUNCTIONS Example 6
E.
f ’(x) > 0 when -2 < x < 0 and f ’(x) < 0 when 0 < x < 2.
Thus, f is increasing on (-2, 0) and decreasing on (0, 2).
2
2'( )
4
xf x
x
DERIVATIVES OF LOG FUNCTIONS Example 6
F. The only critical number is x = 0.
f ’ changes from positive to negative at 0.
Therefore, f(0) = ln 4 is a local maximum by the First Derivative Test.
G.
f ’’(x) < 0 for all x. So, the curve is concave downward on (-2, 2)
and has no inflection point.
DERIVATIVES OF LOG FUNCTIONS Example 62
2 2
2
2 2
(4 )( 2) 2 ( 2 )''( )
(4 )
8 2
(4 )
x x xf x
x
x
x
H. Using that information, we sketch
the curve.
DERIVATIVES OF LOG FUNCTIONS Example 6
© Thomson Higher Education
Find f ’(x) if f(x) = ln |x|.
Since
it follows that:
Thus, f ’(x) = 1/x for all x ≠ 0.
ln if 0( )
ln( ) if 0
x xf x
x x
1if 0
( )1 1
( 1) if 0
xxf x
xx x
DERIVATIVES OF LOG FUNCTIONS Example 7
The result of Example 7 is worth
remembering:
1ln
dx
dx x
DERIVATIVES OF LOG FUNCTIONS Equation 3
The corresponding integration formula
is:
Notice that this fills the gap in the rule for integrating power functions:
The missing case (n = -1) is supplied by Formula 4.
DERIVATIVES OF LOG FUNCTIONS Formula 4
1ln | |dx x C
x
1
11
nn xx dx C n
n
if
Find, correct to three decimal places,
the area of the region under the hyperbola
xy = 1 from x = 1 to x = 2.
DERIVATIVES OF LOG FUNCTIONS Example 8
© Thomson Higher Education
?
Using Formula 4 (without the absolute value
sign, since x > 0), we see the area is:
DERIVATIVES OF LOG FUNCTIONS Example 8
2
2
1 1
1ln
ln 2 ln1
ln 2
0.693
A dx xx
© Thomson Higher Education
Evaluate:
We make the substitution u = x2 + 1 because the differential du = 2x dx occurs (except for the constant factor 2).
Example 9DERIVATIVES OF LOG FUNCTIONS
2 1
xdx
x
Thus, x dx = ½ du and
Notice that we removed the absolute value signs because x2 + 1 > 0 for all x.
DERIVATIVES OF LOG FUNCTIONS Example 9
1 12 22
212
212
ln | |1
ln | 1|
ln ( 1)
x dudx u C
x u
x C
x C
We could use the properties of
logarithms to write the answer as:
However, this isn’t necessary.
2ln 1x C
DERIVATIVES OF LOG FUNCTIONS Example 9
Calculate:
We let u = ln x because its differential du = dx/x occurs in the integral.
When x = 1, u = ln 1 = 0; when x = e, u = ln e = 1. Thus,
Example 10DERIVATIVES OF LOG FUNCTIONS
1
lne xdx
x
121
1 00
ln 1
2 2
e x udx u du
x
The function f(x) = (ln x)/x in Example 10
is positive for x >1.
Thus, the integral represents the area of the shaded region in the figure.
DERIVATIVES OF LOG FUNCTIONS
Calculate:
First, we write tangent in terms of sine and cosine:
This suggests that we should substitute u = cos x since, then, du = -sin x dx and so sin x dx = -du.
DERIVATIVES OF LOG FUNCTIONS Example 11
tan x dx
sintan
cos
xx dx dx
x
Thus,
DERIVATIVES OF LOG FUNCTIONS Example 11
sintan
cosln | |
ln | cos |
x duxdx dx
x uu C
x C
Since
the result of Example 11 can also be
written as:
DERIVATIVES OF LOG FUNCTIONS
ln | cos | ln(1/ | cos |) ln | sec |x x x
tan ln | sec |x dx x C
Formula 5
Formula 7 in Section 7.3 expresses
a logarithmic function with base a in terms
of the natural logarithmic function:
GENERAL LOG FUNCTIONS
lnlog
lna
xx
a
Since ln a is a constant, we can differentiate
as follows:
ln 1 1(log ) (ln )
ln ln lna
d d x dx x
dx dx a a dx x a
GENERAL LOG FUNCTIONS
1(log )
lna
dx
dx x a
Formula 6
Using Formula 6 and the Chain Rule,
we get:
GENERAL LOG FUNCTIONS Example 12
10log (2 sin )
1(2 sin )
(2 sin ) ln10
cos
(2 sin ) ln10
dx
dxd
xx dx
x
x
From Formula 6, we see one of the main
reasons that natural logarithms (logarithms
with base e) are used in calculus:
The differentiation formula is simplest when a = e because ln e = 1.
GENERAL LOG FUNCTIONS
In Section 7.2, we showed that the derivative
of the general exponential function
f(x) = ax, a > 0
is a constant multiple of itself:
EXPONENTIAL FUNCTIONS WITH BASE a
0
1'( ) '(0) '(0) lim
hx
h
af x f a f
h
where
We are now in a position to show that
the value of the constant is f ’(0) = ln a.
EXP. FUNCTIONS WITH BASE a Formula 7
( ) lnx xda a a
dx
We use the fact that e ln a = a:
EXP. FUNCTIONS WITH BASE a
ln (ln )
ln
( ) ( ) (ln )
( ) (ln )
ln
x a x a x
a x
x
d d da e e a x
dx dx dx
e a
a a
Formula 7—Proof
In Example 6 in Section 3.7, we considered
a population of bacteria cells that doubles
every hour.
We saw that the population after t hours is:
n = n02t
where n0 is the initial population.
EXP. FUNCTIONS WITH BASE a
Formula 7 enables us to find
the growth rate:
EXP. FUNCTIONS WITH BASE a
0 2 ln 2tdnn
dt
Combining Formula 7 with the Chain
Rule, we have:
EXP. FUNCTIONS WITH BASE a Example 13
2 2 2
2
(10 ) 10 (ln10) ( )
(2 ln10) 10
d dx x x
dx dx
x x
The integration formula that follows
from Formula 7 is:
EXP. FUNCTIONS WITH BASE a Example 13
1ln
xx aa dx C a
a
EXP. FUNCTIONS WITH BASE a Example 14
55
00
5 0
22
ln 2
2 2
ln 2 ln 231
ln 2
xxdx
The calculation of derivatives of
complicated functions involving products,
quotients, or powers can often be simplified
by taking logarithms.
The method used in the following example is called logarithmic differentiation.
LOGARITHMIC DIFFERENTIATION
Differentiate:
We take logarithms of both sides and use the Laws of Logarithms to simplify:
3 / 4 2
5
1
(3 2)
x xy
x
23 14 2ln ln ln( 1) 5ln(3 2)y x x x
LOGARITHMIC DIFFERENTIATION Example 15
Differentiating implicitly with respect to x gives:
Solving for dy/dx, we get:
2
1 3 1 1 2 35
4 2 3 21
dy x
y dx x xx
2
3 15
4 1 3 2
dy xy
dx x x x
LOGARITHMIC DIFFERENTIATION Example 15
Since we have an explicit expression for y, we can substitute and write:
3 / 4 2
5 2
1 3 15
4 3 2(3 2) 1
dy x x x
dx x xx x
LOGARITHMIC DIFFERENTIATION Example 15
If we hadn’t used logarithmic differentiation
in Example 15, we would have had to use
both the Quotient Rule and the Product Rule.
The resulting calculation would have been horrendous.
LOGARITHMIC DIFFERENTIATION Note
1. Take natural logarithms of both sides
of an equation y = f(x) and use the Laws
of Logarithms to simplify.
2. Differentiate implicitly with respect to x.
3. Solve the resulting equation for y’.
LOGARITHMIC DIFFERENTIATION—STEPS
If f(x) < 0 for some values of x, then ln f(x)
is not defined.
However, we can write | y | = | f(x) | and
use Equation 3.
We illustrate this procedure by proving the general version of the Power Rule—as promised in Section 3.3.
LOGARITHMIC DIFFERENTIATION
If n is any real number and f(x) = xn,
then
Let y = xn and use logarithmic differentiation:
Thus,
Hence,
1'( ) nf x nx
THE POWER RULE
ln ln ln 0n
y x n x x 'y n
y x
1'n
ny xy n n nx
x x
Proof
You should distinguish carefully
between:
The Power Rule [(d/dx) xn = nxn-1], where the base is variable and the exponent is constant.
The rule for differentiating exponential functions [(d/dx) ax = ax ln a], where the base is constant and the exponent is variable
LOGARITHMIC DIFFERENTIATION Note
In general, there are four cases for
exponents and bases:
1
( ) ( )
( )
1. ( ) 0 and areconstants
2. ( ) ( ) '( )
3. (ln ) '( )
4. To find ( / [ ( )] , logarithmic
differentiation can be used, as in the next example.
b
b b
g x g x
g x
da a b
dxdf x b f x f x
dxda a a g x
dx
d dx f x
LOGARITHMIC DIFFERENTIATION
Differentiate .
Using logarithmic differentiation, we have:
ln ln ln
' 1 1(ln )
2
1 ln 2 ln'
2 2
x
x
y x x x
yx x
y x x
x xy y x
x x x
LOGARITHMIC DIFFERENTIATION E. g. 16—Solution 1
xy x
Another method is to write:
ln ln( ) ( ) ( ln )
2 ln
2
x x x x x
x
d d dx e e x x
dx dx dx
xx
x
LOGARITHMIC DIFFERENTIATION E. g. 16—Solution 2
ln( )x x xx e
We have shown that, if f(x) = ln x,
then f ’(x) = 1/x.
Thus, f ’(1) = 1.
Now, we use this fact to express the number e as a limit.
THE NUMBER e AS A LIMIT
From the definition of a derivative as a limit,
we have:
0 0
0
0
1
0
(1 ) (1) (1 ) (1)'(1) lim lim
ln(1 ) ln1lim
1lim ln(1 )
lim ln(1 )
h x
x
x
x
x
f h f f x ff
h xx
x
xx
x
THE NUMBER e AS A LIMIT
As f ’(1) = 1, we have:
Then, by Theorem 8 in Section 2.5 and the continuity of the exponential function, we have:
1
0lim ln(1 ) 1x
xx
1/1
0lim ln(1 )1 ln(1 ) 1
0 0lim lim(1 )
xx
xx x x
x xe e e e x
THE NUMBER e AS A LIMIT
1
0lim(1 ) x
xe x
Formula 8
Formula 8 is illustrated by:
The graph of the function y = (1 + x)1/x.
A table of values for small values of x.
THE NUMBER e AS A LIMIT
If we put n = 1/x in Formula 5, then
n → ∞ as x → 0+.
So, an alternative expression for e is:
1lim 1
n
ne
n
THE NUMBER e AS A LIMIT Formula 9