3.2 The Natural Logarithm Function
3.2
Brian E. Veitch
The Natural Logarithm Function
Definition 3.3 (The Natural Logarithm). The natural logarithmic function is the function
defined by
x
Z
ln x =
1
So ln x is the area under the curve f (x) =
1
dt
t
1
from 1 to x.
t
For 0 < x < 1:
72
3.2 The Natural Logarithm Function
Brian E. Veitch
This means the value of ln x, x < 1, is the the negative of the area shown above. This
should make sense from one of our integration properties.
Z
ln x =
1
x
1
dt = −
t
Z
1
x
1
dt < 0
t
Derivative of the Natural Logarithm
d
1
(ln x) =
dx
x
Proof. It comes directly from the definition of ln x and the Fundamental Theorem of Calc
Part 1.
d
d
ln x =
dx
dx
3.2.1
Z
1
Laws of Logarithms
1. ln(xy) = ln x + ln y
x
2. ln
= ln x − ln y
y
1
3. ln(
= − ln(x)
x
4. ln(xr ) = r ln x
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x
1
1
dt =
t
x
3.2 The Natural Logarithm Function
Example 3.7. Expand ln
Brian E. Veitch
3x2
(x + 1)3
ln
3x2
= ln 3x2 − ln(x + 1)3
3
(x + 1)
= ln 3x2 − 3 ln(x + 1)
p
(x2 + 5)10 cos(x)
Example 3.8. Expand ln
x3
p
p
(x2 + 5)10 cos(x)
2
10
cos(x) − ln x3
ln
=
ln(x
+
5)
+
ln
x3
1
= 10 ln(x2 + 5) + ln cos(x) − 3 ln x
2
Example 3.9. Express as one ln:
1
ln(1 + x2 ) − ln x − 2 ln sin(x)
3
1
(1 + x2 )1/3
ln(1 + x2 ) − ln x − 2 ln sin(x) = ln
3
x sin2 (x)
3.2.2
Derivative of the Natural Logarithm
d
1 du
(ln u) = ·
dx
u dx
d
1
[ln(g(x))] =
· g 0 (x)
dx
g(x)
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3.2 The Natural Logarithm Function
Example 3.10. Find
Brian E. Veitch
d
ln(tan x):
dx
d
1
d
ln(tan x) =
·
[tan(x)]
dx
tan(x) dx
1
· sec2 (x)
=
tan(x)
d
Example 3.11. Find
ln
dx
x+1
√
x−2
You can do this in two ways. We’ll do both though. The first way, we just differentiate
using our current log derivative rule.
1. Take the derivative directly using the chain rule.
1
x+1
d
x+1
d
√
=
ln √
·
x + 1 dx
dx
x−2
x−2
√
x−2
√

1
√
−1/2
x − 2  x − 2 · 1 − (x + 1) · 2 (x − 2)

=
·

x+1
(x − 2)
1
(x − 2) − (x + 1)
2
=
(x + 1)(x − 2)
1
x − 32
2
=
(x + 1)(x − 2)
2. Rewrite ln
x+1
√
x−2
as ln(x + 1) −
1
ln(x − 2)
2
Now differentiate,
d
ln
dx
x+1
√
x−2
75
=
1
1
−
x + 1 2(x − 2)
3.2 The Natural Logarithm Function
Brian E. Veitch
Which one do you think was easier?
Example 3.12. Find
d
(ln(tan(x))2
dx
y 0 = 2 (ln tan(x)) ·
1
· sec2 (x)
tan(x)
1
d
2
x ln
Example 3.13. Find
dx
x2
Use the product rule
d
1
2
1
2
2
2
x ln
=
x
·
x
·
−
+
2x
·
ln
dx
x2
x3
x2
1
= −2x + 2x · ln
x2
= −2x(1 − 2 ln(x))
How could you make this problem a bit easier? You could rewrite the original problem.
2
Note, x ln
1
x2
= −2x2 ln(x), which is far easier to differentiate.
−2x2 ·
1
+ (−4x) ln(x) = −2x − 4x ln(x) = −2x(1 − 2 ln(x))
x
Example 3.14. Given f (x) = ln |x|, show f 0 (x) =
76
1
.
x
3.2 The Natural Logarithm Function
ln |x| =
Brian E. Veitch


ln x,
x>0

ln(−x), x < 0
After you differentiate the piece-wise function, you get

1

 ,
x>0
f 0 (x) = x

 1 · (−1) = 1 , x < 0
−x
x
So what does this mean?
1.
d
1
ln |x| =
dx
x
Z
2.
3.2.3
1
dx = ln |x| + C
x
Logarithmic Differentiation
We use this method when we need to take derivatives of complicated products or quotients.
√
x x2 + 1
Example 3.15. Differentiate y =
(x + 1)2/3
1. Take ln of both sides.
ln y = ln
!
√
x x2 + 1
(x + 1)2/3
= ln x +
1
2
ln(x2 + 1) − ln(x + 1)
2
3
77
3.2 The Natural Logarithm Function
Brian E. Veitch
This looks a bit nicer, right?
2. Differentiate both sides with respect to x
1
y
1
y
dy
1
1
2
=
+
· 2x −
2
dx
x 2(x + 1)
3(x + 1)
dy
1
x
2
·
=
+ 2
−
dx
x x + 1 3(x + 1)
·
Beats doing the quotient rule with a product rule and a chain rule!
3. Solve for
dy
dx
1
x
2
+
−
x x2 + 1 3(x + 1)
√
1
x
2
x x2 + 1
·
+
−
=
(x + 1)2/3
x x2 + 1 3(x + 1)
dy
= y
dx
I dare you to try this without logarithmic differentiation.
Let’s move on to integration using ln x.
Z
Since we know
1
dx = ln |x|, let’s do some examples.
x
Z
Example 3.16. Find
2x
dx
−5
x2
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3.2 The Natural Logarithm Function
Brian E. Veitch
1. Let u = x2 − 5
2. du = 2x dx
3. Make the substitution
Z
Z
2x
1
dx =
du
2
x −5
u
= ln |u| + C
= ln(x2 − 5) + C
Z
4 cos(x)
dx
3 + 2 sin(x)
Example 3.17. Find
1. Let u = 3 + 2 sin(x)
2. du = 2 cos(x) dx → 2du = 4 cos(x) dx
3. Make the substitution
Z
Z
4 cos(x)
2
dx =
du
3 + 2 sin(x)
u
= 2 ln |u| + C
= 2 ln(3 + 2 sin(x)) + C
Z
Example 3.18. Find
tan(x) dx
1. We can’t do much with tan(x). Let’s rewrite tan(x) =
Z
2. So
Z
tan(x) dx =
sin(x)
dx
cos(x)
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sin(x)
cos(x)
3.2 The Natural Logarithm Function
Brian E. Veitch
3. Let u = cos(x)
4. du = − sin(x) dx → −du = sin(x) dx
5. Make the substitution
Z
Z
sin(x)
dx
cos(x)
Z
1
=
(−du)
u
Z
1
=
− du
u
= − ln |u| + C
tan(x) dx =
= − ln | cos(x)| + C
= ln |(cos(x))−1 | + C
= ln | sec(x)| + C
Z
Example 3.19. Evaluate
1
2
4 + u2
du
u3
This problem doesn’t have any trick to it. We just need to break it up into two fractions.
1. Break it up into two separate fractions
Z
1
2
4 + u2
du =
u3
Z
1
2
4
1
+ du
3
u
u
2. At this point, just integrate like you normally would.
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3.2 The Natural Logarithm Function
Z
2
1
Brian E. Veitch
4
1
+ du =
3
u
u
Z
2
1
du
u
1
2
= −2u−2 + ln |u|1
−2
−2
=
+ ln(2) −
+ ln(1)
22
11
3
+ ln(2)
=
2
Z
Example 3.20. Evaluate
e
6
4u−3 +
dx
x ln x
1. This is a substitution problem.
2. Let u = ln(x).
3. Why u = ln(x)? Because du =
1
dx and that’s also in the integral.
x
4. Since it’s a definite integral, we need to change the limits of integration
If x = e, u = ln(e) = 1
If x = 6, u = ln(6)
5. Make the substitution
Z
e
6
dx
=
x ln x
Z
1
ln(6)
1
du
u
= ln |u||ln(6)
1
= ln | ln(6)| − ln | ln(1)|
= ln | ln(6)|
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