Natural Logarithms (Sect. 7.2)
I
Definition as an integral.
I
The derivative and properties.
I
The graph of the natural logarithm.
I
Integrals involving logarithms.
I
Logarithmic differentiation.
Definition as an integral
Recall:
(a) The derivative of y = x n is y 0 = n x (n−1) , for n integer.
Z
x (n+1)
n
n
(b) The integral of y = x is
x dx =
, for n 6= −1.
(n + 1)
Z
dx
(c) Case n = −1:
is neither rational nor trigonometric
x
function. This is a new function.
Definition
The natural logarithm is the
function
Z x
dt
ln(x) =
,
x ∈ (0, ∞).
t
1
In particular: ln(1) = 0.
y
y = ln (x)
ln (x)
y = 1/x
1
+
1
x
x
Definition as an integral
y
Definition
The natural logarithm is the
function
Z x
dt
,
x ∈ (0, ∞).
ln(x) =
1 t
y = ln (x)
ln (x)
y = 1/x
1
+
x
1
x
In particular: ln(1) = 0.
Definition
The number e is the number
satisfying ln(e) = 1, that is,
Z e
dt
= 1.
1 t
(e = 2.718281...).
Natural Logarithms (Sect. 7.2)
I
Definition as an integral.
I
The derivative and properties.
I
The graph of the natural logarithm.
I
Integrals involving logarithms.
I
Logarithmic differentiation.
y
y = ln (x)
1
1
e
x
The derivative and properties
Theorem (Derivative of ln)
The Fundamental Theorem of Calculus implies ln0 (x) =
Proof:
Z
ln(x) =
1
x
dt
t
ln0 (x) =
⇒
1
.
x
1
.
x
Theorem (Chain rule)
0 u 0
For every differentiable function u holds ln(u) = .
u
Proof:
d ln
du
1
d ln(u)
(u)
= u0
=
du
dx
u
dx
⇒
d ln(u)
u 0 (x)
(x) =
.
dx
u(x)
The derivative and properties
Example
Find the derivative of y (x) = ln(3x), and z(x) = ln 2x 2 + cos(x) .
Solution: We use the chain rule.
y 0 (x) =
1
1
(3) =
(3x)
x
⇒
y 0 (x) =
1
.
x
We also use chain rule,
z 0 (x) =
1
(4x − sin(x))
2x 2 + cos(x)
z 0 (x) =
4x − sin(x)
.
2x 2 + cos(x)
Remark: y (x) = ln(3x), satisfies y 0 (x) = ln0 (x).
C
The derivative and properties
Theorem (Algebraic properties)
For every positive real numbers a and b holds,
(a) ln(ab) = ln(a) + ln(b),
a
(b) ln
= ln(a) − ln(b),
b
1
(c) ln
= − ln(a),
a
(d) ln(ab ) = b ln(a),
(product rule);
(quotient rule);
(reciprocal rule);
(power rule).
Proof of (a): (only)
1
1
a = = ln0 (x)
x
ax
Therefore ln(ax) = ln(x) + c. Evaluating at x = 1 we obtain c.
The function y (x) = ln(ax) satisfies y 0 (x) =
ln(a) = ln(1) + c ⇒ c = ln(a) ⇒ ln(ax) = ln(x) + ln(a).
The derivative and properties
Example
h (x + 1)2 i
Compute the derivative of y (x) = ln
.
3(x + 2)
Solution: Before computing the derivative of y , we simplify it,
y = ln (x + 1)2 − ln 3(x + 2) ,
y = 2 ln(x + 1) − ln(3) + ln(x + 2) .
The derivative of function y is: y 0 = 2
y0 =
2(x + 2) − (x + 1)
(x + 1)(x + 2)
⇒
1
1
−
.
(x + 1) (x + 2)
y0 =
(x + 3)
.
(x + 1)(x + 2)
C
Natural Logarithms (Sect. 7.2)
I
Definition as an integral.
I
The derivative and properties.
I
The graph of the natural logarithm.
I
Integrals involving logarithms.
I
Logarithmic differentiation.
8
The graph of the natural logarithm
y
y = ln (x)
Remarks:
y
(a) A vertical asymptote at
x = 0.
y
1
0
e
x
x
8
(b) No horizontal asymptote.
x
1
8
The graph of ln function has:
Proof: Recall e = 2.718281... > 1 and ln(e) = 1.
(a): If x = e n , then ln(e n ) = n ln(e) = n. Hence
lim ln(x) = ∞.
x→∞
1
1
(b): If x = n , then ln n = − ln(e n ) − n ln(e) = −n. Hence
e
e
lim+ ln(x) = −∞.
x→0
Natural Logarithms (Sect. 7.2)
I
Definition as an integral.
I
The derivative and properties.
I
The graph of the natural logarithm.
I
Integrals involving logarithms.
I
Logarithmic differentiation.
Integrals involving logarithms.
Z
dx
= ln(|x|) + c for x 6= 0 and c ∈ R.
x
Indeed, for x > 0 this is the definition of logarithm.
And for x < 0, we have that −x > 0, then,
Z
Z
dx
(−dx)
= ln(−x) + c,
−x > 0.
=
x
(−x)
Remark: It holds
We conclude,
Z
dx
=
x
Z
Remark: It also holds
(
ln(−x) + c
if x < 0,
ln(x) + c
if x > 0.
f 0 (x)
dx = ln(|f (x)|) + c, for f (x) 6= 0.
f (x)
Integrals involving logarithms.
Remarks:
Z
tan(x) dx = − ln(| cos(x)|) + c. Indeed,
(a)
Z
Z
sin(x)
dx u = cos(x), du = − sin(x) dx.
cos(x)
Z
Z
du
tan(x) dx = −
= − ln(|u|) + c = − ln(| cos(x)|) + c.
u
Z
(b)
cot(x) dx = ln(| sin(x)|) + c. Indeed,
tan(x) dx =
Z
Z
cos(x)
dx u = sin(x), du = cos(x) dx.
sin(x)
Z
Z
du
cot(x) dx =
= ln(|u|) + c = ln(| sin(x)|) + c.
u
cot(x) dx =
Integrals involving logarithms.
Example
Z
Find y (t) =
3 sin(t)
dt.
(2 + cos(t))
Solution:
Z
y (t) =
3 sin(t)
dt,
(2 + cos(t))
Z
y (t) =
u = 2 + cos(t),
3(−du)
= −3
u
Z
du = − sin(t) dt.
du
= −3 ln(|u|) + c
u
We conclude that y (t) = −3 ln |2 + cos(t)| + c.
C
Natural Logarithms (Sect. 7.2)
I
Definition as an integral.
I
The derivative and properties.
I
The graph of the natural logarithm.
I
Integrals involving logarithms.
I
Logarithmic differentiation.
Logarithmic differentiation
Remark: Logarithms can be used to simplify the derivative of
complicated functions.
Example
x 3 (x + 2)2
Find the derivative of y (x) =
.
cos3 (x)
h x 3 (x + 2)2 i
Solution: First compute ln[y (x)] = ln
,
cos3 (x)
ln[y (x)] = ln x 3 (x + 2)2 − ln cos3 (x) ,
ln[y (x)] = ln x 3 + ln (x + 2)2 − ln cos3 (x) ,
ln[y (x)] = 3 ln(x) + 2 ln(x + 2) − 3 ln cos(x) .
Logarithmic differentiation
Example
x 3 (x + 2)2
Find the derivative of y (x) =
.
cos3 (x)
Solution: Recall: ln[y (x)] = 3 ln(x) + 2 ln(x + 2) − 3 ln cos(x) .
y 0 (x)
3
2
3 sin(x)
= +
+
.
y (x)
x
(x + 2)
cos(x)
h3
2
3 sin(x) i
y (x) =
+
+
y (x).
x
(x + 2)
cos(x)
0
We conclude that
2
3 sin(x) i cos3 (x)
y (x) =
+
+
.
x
(x + 2)
cos(x) x 3 (x + 2)2
0
h3
C