Lecture 3Section 7.3 The Logarithm Function, Part II
Jiwen He
Section 7.2: Highlights
1
2
Properties of the Log Function
Z x
1
d
1
• ln x =
dt,
(ln x) = > 0.
t
dx
x
1
• ln 1 = 0,
ln e = 1.
• ln(xy) = ln x + ln y, ln(x/y) = ln x − ln y.
• ln xr = r ln x, ln er = r.
• domain = (0, ∞),
range = (−∞, ∞).
• lim ln x = −∞,
lim ln x = ∞.
x→∞
x→0+
• lim
x→∞
ln x
= 0,
xr
lim xr ln x = 0.
x→0+
Limits
1
Differentiation and Graphing
1.1
Chain Rule
Differentiation: Chain Rule
Theorem 1.
1 d
d
ln u(x) =
u(x) ,
dx
u(x) dx
for x s.t. u(x) > 0.
du
d
d
1 du
ln u =
ln u
=
.
dx
du
dx
u dx
d
1
d
1
2x
•
ln(1 + x2 ) =
1 + x2 =
· 2x =
,
dx
1 + x2 dx
1 + x2
1 + x2
2
⇐ 1 + x > 0.
Proof. By the chain rule,
Examples 2.
for all x
•
d
1
d
1
3
ln(1 + 3x) =
1 + 3x =
·3 =
, for all x > − 13
dx
1 + 3x dx
1 + 3x
1 + 3x
⇐ 1 + 3x > 0.
Example
p
Example 3. Find the domain of f and find f 0 (x) if f (x) = ln x 4 + x2 .
Solution
√
• For x ∈ domain(f ), we need x 4 + x2 > 0, thus x > 0.
3
• Before differentiating f , simplify it:
p
p
1
f (x) = ln x 4 + x2 = ln x + ln 4 + x2 = ln x + ln 4 + x2 .
2
• Thus
f 0 (x) =
1.2
0
1 1 1
1 1 1
1
x
4 + x2 = +
+
· 2x = +
.
2
2
x 24+x
x 24+x
x 4 + x2
Graphing
Example
4
x4
Example 4. Let f (x) = ln
. Specify the domain of f . On what
x−1
intervals does f increase? Decrease? Find the extrem values of f .Determine the
concavity and inflection points. Skectch the graph, specifying the asymptotes.
Solution
x4
> 0, we need x > 1, thus
For
x−1
domain(f ) = (1, ∞).
f 0 (x) =
• Simplify f (x) = 4 ln x−ln (x − 1). Then
1
3x − 4
4
−
=
.
x x−1
x(x − 1)
• Thus f ↓ on (1, 43 ) and ↑ on ( 43 , ∞).
• At x = 43 , f 0 (x) = 0. Thus
is the (only) local and absolute minimum.
f ( 34 ) = 4 ln 4 − 3 ln 3 ≈ 2.25
f 00 (x) = −
1
• From f 0 (x) = x4 − x−1
, we have
1
(x − 2)(3x − 2)
4
+
=−
.
x2
(x − 1)2
x2 (x − 1)2
/ domain(f ) is ignored). Then, the graph is
• At x = 2, f 00 (x) = 0 ( 23 ∈
concave up on (1, 2), concave down on (2, ∞).
• The point
is the only point of inflection.
• From f 0 (x) =
4
x
−
1
x−1 ,
(2, f (2)) = (2, 4 ln 2) ≈ (2, 2.77)
we have
lim f 0 (x) = −∞,
lim f 0 (x) = 0.
x→∞
x→1+
• From f (x) = 4 ln x − ln(x − 1), we have
lim f (x) = ∞,
x→1+
• The line x = 1 is a vertical asymptote.
5
lim f (x) = ∞.
x→∞
Example
Example 5. Let f (x) = x ln x. Specify the domain of f and find the intercepts. On what intervals does f increase? Decrease? Find the extrem values of
f .Determine the concavity and inflection points. Skectch the graph.
Solution
• ln x is defined only for x > 0, thus
domain(f ) = (0, ∞).
6
• There is no y-intercept. Since
f (1) = 1 · ln 1 = 0,
x = 1 is the only x-intercept.
• We have f 0 (x) = x ·
1
+ ln x = 1 + ln x.
x
• For f 0 (x) = 0, we have
1 + ln x = 0 ⇒ ln x = −1 ⇒ x =
Thus f ↓ on (0, 1e ) and ↑ on ( 1e , ∞).
• At x = 1e , f 0 (x) = 0. Thus
is the (only) local and absolute minimum.
f ( 1e ) =
• From f 0 (x) = 1+ln x, we have
f 00 (x) =
1
e
ln 1e = − 1e ≈ −0.368.
1
> 0,
x
• Then, the graph is concave up on (0, ∞).
• There is no point of inflection.
• From f 0 (x) = 1 + ln x, we have
lim f 0 (x) = −∞,
x→0+
lim f 0 (x) = ∞.
x→∞
• From f (x) = x ln x, we have
lim f (x) = 0,
x→0+
lim f (x) = ∞.
x→∞
Quiz
Quiz
2
2.1
1.
ln 1 =? :
(a) −1,
(b) 0,
(c) 1.
2.
ln e =? :
(a) 0,
(b) 1,
(c) e.
ln |x|
Properties
f (x) = ln |x|, x 6= 0
7
1
.
e
for all x > 0.
Graph
The graph has two branches:
each is the mirror image of the other.
Theorem 6.
1
d
ln |x| =
dx
x
y = ln(−x), x < 0 and y = ln x, x > 0,
Z
⇔
1
dx = ln |x| + C
x
1
d
d
ln |x| =
ln x = .
dx
dx
x
d
1 d
1
d
ln |x| =
ln(−x) =
(−x) = .
• For x < 0,
dx
dx
−x dx
x
Proof.
• For x > 0,
Z
Power Rule:
xn dx
8
Power Rule
Z
xn dx =



1
xn+1 + C,
n+1
if n 6= −1,


ln |x| + C,
if n = −1.
Example 7.
Z
2.2
x+1
dx =
x2
Z 1
1
+
x x2
dx = ln |x| −
1
+ C.
x
Chain Rule
Differentiation: Chain Rule
Theorem 8.
1 d
d
ln |u(x)| =
u(x) ,
dx
u(x) dx
for x s.t. u(x) 6= 0.
du
d
1 du
d
ln |u| =
ln |u|
=
.
dx
du
dx
u dx
1
d
−3x2
d
ln |1 − x3 | =
1 − x3 =
.
Examples 9.
•
3
dx
1 − x dx
1 − x3
x − 1
d
= d (ln |x − 1|)− d (ln |x − 2|)
ln •
dx
x − 2
dx
dx
1
.
x−2
Proof. By the chain rule,
2.3
Logarithmic Differentiation
Logarithmic Differentiation
Theorem 10. Let g(x) = g1 (x) · g2 (x) · · · gn (x). Then
0
gn0 (x)
g1 (x) g20 (x)
0
+
+ ··· +
g (x) = g(x)
.
g1 (x) g2 (x)
gn (x)
Proof.
• First write
ln |g(x)| = ln (|g1 (x)| · |g2 (x)| · · · |gn (x)|)
= ln |g1 (x)| + ln |g2 (x)| + · · · + ln |gn (x)|.
• Then differentiate
g 0 (x)
g 0 (x) g20 (x)
g 0 (x)
= 1
+
+ ··· + n
.
g(x)
g1 (x) g2 (x)
gn (x)
• Multiplying by g(x) gives the result.
9
=
1
−
x−1
Examples
Examples 11. Find f 0 (x) if
• g(x) = x(x − 1)(x − 2)(x − 3).
• g(x) =
(x2 + 1)3 (2x − 5)2
.
(x2 + 5)2
Solution
ln |g(x)| = ln |x| + ln |x − 1| + ln |x − 2| + ln |x − 3|.
g 0 (x)
1
1
1
1
= +
+
+
.
g(x)
x x−1 x−2 x−3
1
1
1
1
0
+
+
+
g (x) = x(x − 1)(x − 2)(x − 3)
.
x x−1 x−2 x−3
ln |g(x)| = 3 ln |x2 + 1| + 2 ln |2x − 5| − 2 ln |x2 + 5|.
2x
2
2x
g 0 (x)
=3 2
+2
−2 2
.
g(x)
x +1
2x − 5
x +5
6x
4
(x2 + 1)3 (2x − 5)2
4x
0
+
g (x) =
−
.
(x2 + 5)2
x2 + 1 2x − 5 x2 + 5
Quiz (cont.)
Quiz (cont.)
3.
4.
3
3.1
lim ln x =? :
(a) −∞,
(b) 0,
(c) ∞.
lim ln x =? :
(a) −∞,
(b) 0,
(c) ∞.
x→0+
x→∞
Integration and Trigonometric Functions
u-Substitution
Integration: u-Substitution
Theorem 12.
Z 0
g (x)
dx = ln |g(x)| + C,
g(x)
10
x 6= 0.
Proof.
Let u = g(x), thus du = g 0 (x)dx, then
Z 0
Z
g (x)
1
dx =
du = ln |u| + C = ln |g(x)| + C.
g(x)
u
Z
x2
Example 13. Calculate
dx. Let u = 1 − 4x3 , thus du = −12x2 dx,
3
1
−
4x
Z
Z
x2
1 1
1
1
then
dx
=
−
du = − ln |u| + C = − ln |1 − 4x3 | + C.
3
1 − 4x
12 u
12
12
Examples: u-Substitution
Z
ln x
Examples 14.
•
dx.
x
Z
1
√
√ dx.
•
x(1 + x)
Z 2
6x2 + 2
dx.
•
3
1 x +x+1
Solution
1
Set u = ln x, du = dx. Then
x
Z
Z
ln x
1
1
dx = udu = u2 + C = (ln x)2 + C.
x
2
2
√
1
du = √ dx. Then
2 x
Z
Z
√ 1
1
√
√ dx = 2
du = 2 ln |u| + C = 2 ln 1 + x + C.
u
x(1 + x)
Set u = 1 +
x,
Set u = x3 + x + 1, du = (3x2 + 1)dx. At x = 1, u = 3; at x = 2, u = 11.
Then
Z 11
Z 2
6x2 + 2
1
11
dx = 2
du = 2 [ln |u|]3 = 2 (ln 11 − ln 3) .
3
u
1 x +x+1
3
• Natural log arises (only) when integrating a quotient whose numerator is
the derivative of its denominator (or a constant multiple of it).
Z 0
Z
g (x)
1
dx =
du = ln |u| + C = ln |g(x)| + C.
g(x)
u
11
3.2
Trigonometric Functions
Integration of Trigonometric
Functions
Z
d
Recall that [1ex]
cos x dx = sin x + C
⇔
sin x = cos x.
dx Z
Z
d
cos x = − sin x. [1ex]
sec2 x dx =
sin x dx = − cos x + C
⇔
dx
Z
d
tan x + C
⇔
tan x = sec2 x.
csc2 x dx = − cot x + C
⇔
dx Z
d
d
cot x = − csc2 x. [1ex]
sec x tan x dx = sec x + C
⇔
sec x =
dx
dx
Z
d
sec x tan x.
csc x cot x dx = − csc x + C ⇔
csc x = − csc x cot x.
dx
New Integration Formulas
Integration
of Trigonometric Functions
Z
Z
tan x dx = − ln | cos x| + C.
cot x dx = ln | sin x| + C.
Z
ln | sec x + tan x| + C.
csc x dx = ln | csc x − cot x| + C.
Z
sec x dx =
Proof.
Set u = cos x, du = − sin x dx, then
Z
Z
Z
sin x
1
tan x dx =
dx = −
du = − ln |u| + C
cos x
u
= − ln | cos x| + C.
Set u = sin x, du = cos x dx, then
Z
Z
Z
1
cos x
dx =
du = ln |u| + C
cot x dx =
sin x
u
= ln | sin x| + C.
Set u = sec x + tan x, du = (sec x tan x + sec2 x) dx, then
Z
Z
Z
sec x + tan x
sec x tan x + sec2 x
sec x dx = sec
dx =
dx
sec x + tan x
sec x + tan x
Z
1
du = ln |u| + C = ln | sec x + tan x| + C.
=
u
Set u = csc x − cot x, du = (− csc x cot x + csc2 x) dx, then
Z
Z
Z
csc x − cot x
− csc x cot x + csc2 x
csc x dx = csc
dx =
dx
csc x − cot x
csc x − cot x
Z
1
du = ln |u| + C = ln | csc x − cot x| + C.
=
u
12
Z
Examples:
du
u
Z
Examples 15.
Z
•
•
sec2 3x
dx.
1 + tan 3x
x sec x2 dx.
Z
tan(ln x)
dx.
x
Solution
Set u = 1 + tan 3x, du = 3 sec2 3x dx:
Z
Z
sec2 3x
1
1
dx =
du
1 + tan 3x
3
u
1
1
= ln |u| + C = ln |1 + tan 3x| + C.
3
3
•
Set u = x2 , du = 2x dx:
Z
Z
1
1
sec u du = ln | sec u + tan u| + C
x sec x2 dx =
2
2
1
2
= ln | sec x + tan x2 | + C.
2
Set u = ln x, du =
Z
1
x
dx:
Z
tan(ln x)
dx = tan u du = ln | sec u| + C
x
= ln | sec(ln x)| + C.
Outline
Contents
1 Differentiation
1.1 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
4
2 ln |x|
2.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Log Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . .
7
7
9
9
3 Integration
10
3.1 u-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . 12
13