Logarithm and Exponential Derivatives and
Integrals
James K. Peterson
Department of Biological Sciences and Department of Mathematical Sciences
Clemson University
September 3, 2013
Outline
Exponential Functions: Integrals and Derivatives
Derivative Examples
Integral Examples
Logarithm Functions: Integrals and Derivatives
Derivative Examples
Integral Examples
Abstract
This lecture is going to talk the derivatives and integrals of natural
logarithm function ln(x) and the exponential function exp(x) = e x .
We know
I
exp is a continuous function of x for all x,
I
limx → ∞ exp(x) = ∞,
I
limx → −∞ exp(x) = 0,
I
exp(0) = 1,
I
( exp(x) )0 = exp(x),
I
If x and y are positive numbers then
exp(x + y ) = exp(x) exp(y ),
I
If x and y are positive numbers then
exp(x − y ) = exp(x) exp(−y )
exp(x)
,
=
exp(y )
I
If x and y are positive numbers then
(exp(x))y
= exp(xy ),
.
We have two new rules:
I
d u(x) = e u(x) u 0 (x)
e
dx Z
e u du = e u + C
I
Let’s do some examples.
Example
Find the derivative of exp(x 2 + x + 1).
Solution
d
exp(x 2 + x + 1) = exp(x 2 + x + 1) (2x + 1)
dx
Example
Find the derivative of exp(tan(x)).
Solution
d
(exp(tan(x))) = exp(tan(x)) sec2 (x)
dx
Example
Find the derivative of exp(exp(x 2 )).
Solution
d
exp(exp(x 2 )) = exp(exp(x 2 )) exp(x 2 ) 2x
dx
Example
Find the derivative of exp(−5x 3 + 2x − 8).
Solution
d
exp(−5x 3 + 2x − 8) = exp(−5x 3 + 2x − 8) (−15x 2 + 2)
dx
Example
Find the derivative of exp(−3x).
Solution
d
(exp(−3x)) = exp(−3x) (−3)
dx
Example
Find the derivative of exp(5x).
Solution
d
(exp(5x)) = exp(5x) 5
dx
Before we go any further, you need to know that historically, many
people got tired of writing exp all the time. They shortened the
notation as follows: exp(u(x)) = e u(x) .
Sometimes it is easier to read a complicated exponential expression
using the e u(x) notation and sometimes not. Judge for yourselves.
1.
d x 2 +x+1 2
e
= e x +x+1 (2x + 1)
dx
2.
d
e −3t = e −3t (−3)
dt
3.
d
e 5t = e 5t 5
dt
Homework 15
15.1 Find the derivative of exp(−3t).
15.2 Find the derivative of exp(t 2 + 2t − 4).
15.3 Find the derivative of e 1/t .
15.4 Find the derivative of e t /(e t + e −t ).
15.5 Find the derivative of (t 2 + 2t − 8) e 5t .
15.6 Find the derivative of e 4t .
15.7 Find the derivative of e −3t .
15.8 Find the derivative of e 18t .
Example
Find the integral of
R
exp(x 2 + x + 1) (2x + 1).
Solution
Z
exp(x 2 + x + 1) (2x + 1) =
Z
exp(u) du,
use substitution u = x 2 + x + 1
= exp(u) + C
= exp(x 2 + x + 1) + C
Example
Find the integral of
R
exp(tan(x)) sec2 (x) dx.
Solution
Z
exp(tan(x)) sec2 (x) dx
Z
=
exp(u) du,
use substitution u = tan(x);
= exp(u) + C
= exp(tan(x)) + C
Example
Find the integral of
R
exp(t 2 ) 3t dt.
Solution
Z
exp(t 2 ) 3t dt =
Z
exp(u)3(du/2),
use substitution
u = t 2;
Z
= (3/2)
exp(u) du
= (3/2) exp(u) + C
= (3/2) exp(t 2 ) + C
Example
Find the integral of
R
exp(−3t) dt.
Solution
Z
Z
exp(u) du/(−3), (u = −3t)
exp(−3t) dt =
use Zsubstitution u = −3t;
1
=
exp(u) du
−3
1
1
= − exp(u) + C = − exp(−3t) + C
3
3
Example
Find the integral of
R
exp(5t) dt.
Solution
Z
Z
exp(5t) dt =
=
=
exp(u) du/(5), (u = 5t)
Z
1
exp(u) du
5
1
1
exp(u) + C =
exp(5t) + C
5
5
You should see the problems above using the other notion also.
Here they are:
1.
Z
2
Z
e t 3t dt =
e u 3(du/2), (u = t 2 )
Z
= (3/2)
e u du
2
= (3/2) e u + C = (3/2) e t + C
2.
Z
Z
e −3t dt =
e u du/(−3), (u = −3t)
Z
1
=
e u du
−3
1
1
= − e u + C = − e −3t + C
3
3
Homework 16
16.1 Find the integral
R
exp(−3t) dt.
16.2 Find the integral
R
exp(t 2 + 2t − 4) (2t + 2)dt.
16.3 Find the integral
R
e t t 2 dt.
16.4 Find the integral
R
t e 5t dt.
16.5 Find the integral
R
e 14t dt.
16.6 Find the integral
R5
16.7 Find the integral
R2
3
2
−1
0
e −33t dt.
e 4t dt.
We know
I
ln is a continuous function of x for positive x,
I
limx → ∞ ln(x) = ∞,
I
limx → 0 ln(x) = −∞,
I
ln(1) = 0,
I
ln(e) = 1,
I
( ln(x) )0 =
1
x,
and we know
I
If x and y are positive numbers then
ln(xy ) = ln(x) + ln(y ),
I
If x and y are positive numbers then
x
ln( ) = ln(x) − ln(y ),
y
I
If x and y are positive numbers then
ln(x y ) = y ln(x).
We can say more about the derivative of the logarithm function.
Then ln( | x | ) is nicely defined at all x not zero.
I
ln(| x |) =
I
ln(x)
if x > 0
ln(−x) if x < 0
If x is negative, using the chain rule we have
d
(ln(−x)) =
dx
=
=
1 d
(−x)
−x dx
1
(−1)
−x
1
x
Thus,
I
d
(ln | x |)) =
dx
=
I
d
dx
d
dx
(ln(x))
if x > 0
(ln(−x)) if x < 0
1
if x is not 0
x
We conclude that
d
(ln(| x |)) =
dx
I
=
1
, if x 6= 0
x
So the antiderivative of 1/x is ln(|x|) + C .
1
x
1
x
if x > 0
if x < 0
We have two new rules:
I
d
1
(ln( u(x) ) =
u 0 (x)
dx
u(x)
Z
1
du = ln(|u|) + C
u
I
Let’s do some examples.
Example
Find the derivative of ln(x 2 + x + 1).
Solution
d
ln(x 2 + x + 1) =
dx
1
(2x + 1)
x2 + x + 1
Example
Find the derivative of ln(tan(x)).
Solution
d
(ln(tan(x))) =
dx
1
sec2 (x)
tan(x)
Now the natural logarithm here is only defined when tan(x) is
positive. We usually just assume that we are only interested in
evaluating the expression ln(tan(x) for such x; i.e. by writing
ln(tan(x), we are tacitly assuming that tan(x) is positive. Another
way of looking at this, is that the domain of the function ln(tan(x)
is the set of x where tan(x) is positive. However, it is best to train
ourselves to think about the restrictions on the domain without
being so explicit!
Example
Find the derivative of ln(5x).
Solution
d
(ln(5x)) =
dx
1
1
5 =
5x
x
Here the implied domain of ln(5x) is all positive x. We can again
use the properties of ln to get this result another way.
d
(ln(5x)) =
dx
=
d
d
(ln(x)) +
(ln(5))
dx
dx
1
x
Example
Find the derivative of ln(5x 2 + 6x + 9).
Solution
d
ln(5x 2 + 6x + 9) =
dx
1
(10x + 6).
5x 2 + 6x + 9
Example
Find the derivative of ln(5x 3 + 6x 2 + 9x − 25).
Solution
d
ln(5x 3 + 6x 2 + 9x − 25) =
dx
15x 2 + 12x + 9
.
5x 3 + 6x 2 + 9x − 25
Homework 17
17.1 Find the derivative of ln(t 2 + 1).
17.2 Find the derivative of ln(t 3 + t 2 + 5).
17.3 Find the derivative of ln(y 2 + 3y − 12).
17.4 Find the derivative of ln(x 4 + 25x 3 − 23x + 100).
17.5 Find the derivative of t ln(t) − t.
Example
Find the integral
R
1
x 2 +x+1
(2x + 1).
Solution
Z
1
(2x + 1) =
x2 + x + 1
Z
1
du,
u
use substitution u = x 2 + x + 1
= ln(| u |) + C
= ln(| x 2 + x + 1 |) + C
Example
Find the integral
R
1
tan(x)
sec2 (x) dx.
Solution
Z
1
sec2 (x) dx
tan(x)
Z
=
1
du,
u
use substitution u = tan(x);
= ln(| u |) + C
= ln(| tan(x) |) + C
Example
Find the integral
R
et
1 + et
dt.
Solution
Z
et
dt =
1 + et
Z
1
du,
u
use substitution u = 1 + e t ; du = e t dt
= ln(| u |) + C
= ln(| 1 + |e t |) + C ,
but 1 + e t is always positive
= ln(1 + e t ) + C
Example
Find the integral
R
5x
4 + x2
dx.
Solution
Z
5x
dx
4 + x2
=
=
=
=
Z
5
1
du,
2
u
use substitution u = 4 + x 2 ;
5
ln(| u |) + C
2
5
ln(| 4 + x 2 |) + C ,
2
but 4 + x 2 is always positive
5
ln(4 + x 2 ) + C
2
Example
Find the integral
R
tan(w )dw .
Solution
Z
Z
tan(w )dw
=
sin(w )
dw
cos(w )
1
(−du),
u
use substitution u = cos(w ); du = − sin(w )dw
Z
=
= − ln(| u |) + C
= − ln(| cos(w ) |) + C
Homework 18
18.1 Find the integral
R
ln(t)
t
18.2 Find the integral
R
2t/(t 2 + 4) dt.
18.3 Find the integral
R
2s 2 /(s 3 + 9) ds.
18.4 Find the integral
R
18.5 Find the integral
R
18.6 Find the integral
dt ( let u = ln(t)).
8z 2 + 3
dz.
8z 3 + 9z + 18
40z 4 + 27z 2 + 36z
8z 5 + 9z 3 + 18z 2 +20
R2
t
0 4t 2 + 3 dt.
dz.