Download Do Now: Aim: How do we differentiate the natural logarithmic function?

Aim: How do we differentiate the natural
logarithmic function?
Do Now:
Evaluate:

5
1
1
dx
x
Power Rule
n 1
x
n
 x dx  n  1  C , n  1
Aim: Differentiating Natural Log Function
Course: Calculus
Inverse Exponential Function
Exponential Equation
Exponential example
y = bx
y = 2x
Exponential Equation
Exponential
example
x
b
Logarithm
=
Exponent
Inverse of
Inverse of
x = by
y=
x = 2y
“x
is
the
logarithm
of
y”
Logarithmic Equation Logarithmic example
y = logbx y = log x y = log2x
b
logb x = y if and only if by = x
“y is the logarithm of x
The expression logb x is read as the
“log base b of x”. The function f(x) = logb x
is the logarithmic function with base b.
Aim: Differentiating Natural Log Function
Course: Calculus
Natural Logarithmic Function
f(x) = logex = ln x,
x>0
1. ln 1 = 0 because e0 = 1
2. ln e = 1
because e1 = e
3. ln ex = x because ex = ex
4. e ln x  x
inverse property
5. If ln x = ln y, then x = y
4
x
1

lim 1    e
x 
x

vx = ex
2
-5
5
The logarithmic
function with base e
ux = ln x
is called the natural
log function.
Aim: Differentiating Natural Log Function
Course: Calculus
-2
-4
Properties of Natural Log
1. The domain is (0, ) and the range is
(-, ).
2. The function is continuous,
increasing, and one-to-one.
3. The graph is concave down.
If a and b are positive numbers and n is
rational, then the following are true
• ln(1) = 0
4
2
• ln(ab) = ln a + ln b
• ln(an) = n ln a
-5
• ln (a/b) = ln a – ln b
Aim: Differentiating Natural Log Function
5
-2
-4
Course: Calculus
ux = ln x
Using Properties of Natural Logarithms
Rewrite each expression:
1
1. ln  ln e 1
e
ln ex = x
= -1
2. ln e2
=2
ln ex = x
3. ln e0
=0
ln ex = x because ex = ex
4. 2ln e
=2
ln e = 1 because e1 = e
Aim: Differentiating Natural Log Function
Course: Calculus
Model Problems
Use natural logarithms to evaluate log4 30
log c M
log b M 
log c b
ln 30 3.401198...
log 4 30 
 `
 2.4538....
ln 4
1.38629...

Given ln 2  0.693, ln 3  1.099, and
ln 7  1.946, use the properties of logs to
approximate a) ln 6
b) ln 7/27
ln 6
= ln (2 • 3) = ln 2 + ln 3  0.693 + 1.099  1.792
ln 7/27 = ln 7 – ln 27 = ln 7 – 3 ln 3  1.946 – 3(1.099) 
-1.351
Aim: Differentiating Natural Log Function
Course: Calculus
Model Problems
Use properties of logarithms to rewrite
3x  5
ln
7
x

ln
2
3

= ln(3x – 5)1/2 – ln 7
= 1/2 ln(3x – 5) – ln 7
2
x x 1
3
2

 ln( x 2  3)2  ln x 3 x 2  1


 2ln( x  3)   ln x  ln x 2  1

2

 2ln( x  3)  ln x  ln x  1
2
2



13
1
 2ln( x  3)  ln x  ln x 2  1
3
2
Aim: Differentiating Natural Log Function
Course: Calculus
13



Power Rule – the exception
 x3 
Dx    x 2 
 3 
 x2 
Dx    x 1 
 2 
Dx  x   x 0 
Dx  ???   x


1
D x  x 1  x 2
3
x
2
x
 dx  3  C
2
x
1
x
 dx  2  C
 dx  x  C
dx
 
 ???  C
x
dx
  2   x 1  C
x
Aim: Differentiating Natural Log Function
Course: Calculus
Power Rule – the exception
Evaluate:

5
1

5
1
1
dx
x
1
x dx 
x
5
5

x 
 
  !!!!
0 1
0 1
1  1
0
Power Rule
n 1
x
n
 x dx  n  1  C , n  1
don’t work!
Aim: Differentiating Natural Log Function
Course: Calculus
Power Rule – the exception
Power Rule
n 1
x
n
 x dx  n  1  C , n  1
no antiderivative for f(x) = 1/x
2nd Fundamental Theorem of Calculus
d  x
dF

f  t dt  f ( x ) 


dx  a
dx
Definition of the Natural Logarithmic Function
accumulation function
1
ln x   dt , x  0
1 t
The domain of the natural logarithmic
function is the set of all positive real
Aim: Differentiating numbers.
Natural Log Function
Course: Calculus
x
4
Definition of the Natural Log Function
2
ln x  
x
1
1
dt ,
t
x0
-5
5
-2
4
3.5
4
1
f x  =
x
3.5
3
3
2.5
2.5
ln x is positive
when x > 1
2
1.5
1.5
1
0.5
0.5
1
2
3
x
4
ux = ln x
-4
ln x is negative
when x < 1
2
1
1
1
f x  =
x
x 1
1
2
3
4
x1
1
If x  1,  dt  0
If 0 < x  1,  dt  0
1 t
1 t
11
ln(1) = 0   dt  0
1 t
The natural
log function
measures
the area
Aim: Differentiating
Natural Log Function
Course: Calculus
under the curve f(x) = 1/x between 1 and x.
x
4
e
ln x  
x
1
1
dt  1
t
2
(e, 1)
-5
5
What is the value of x? e
e1
ln e   dt  1
1 t
4
3.5
1
f x  =
x
-2
-4
ux = ln x
ln e  1
loge e  1  e1  e
3
2.5
2
1
Area =  dt  1
1 t
e
1.5
 
ln e n 
 n ln e
 n  1
1
n
0.5
1
1
2
ex
3
4
Aim: Differentiating Natural Log Function
Course: Calculus
1 Natural Log Function
The Derivative
of
the
ln x   dt , x  0
1 t
2nd Fundamental Theorem of Calculus
d  x
dF

f  t dt  f ( x ) 



a
dx 
dx
x
d  x1  1 d
dt     ln x  x  0


dx  1 t  x dx
Let u be a differentiable function of x .
d
1
ln x   , x  0

dx
x
Chain Rule
d
1 du u '
ln u 
 ,u0

dx
u dx u
Aim: Differentiating Natural Log Function
Course: Calculus
Model Problems
d
 ln  2 x  
1.
dx
u  2x
u’ = 2
d
u'
2
1
 ln  2 x   


dx
u 2x x
d 

2.
ln
x

dx 
u
x
1 1 2
u'  x
2
d  1 1
1 1 1 2
1
ln    u '  1 2  x


dx  u  u
x
2
2x
Aim: Differentiating Natural Log Function
Course: Calculus
Model Problems


d 
2

ln
x

1

dx 
u  x2  1
u’ = 2x
u'
2x

 2
u x 1
d
d
d




x ln x   x   ln x     ln x    x  

dx
 dx

 dx

1
 x     ln x  1  1ln x
 x
d 
3
2 d

 ln x    3  ln x   ln x 

dx
dx
2 1
 3  ln x 
x
Aim: Differentiating Natural Log Function
Course: Calculus
Rewrite Before Differentiating
Differentiate f ( x )  ln x  1
rewrite
f ( x )  ln  x  1
12
1
 ln  x  1
2
d 1
 1 d
 ln  x  1 
ln  x  1  

dx  2
 2 dx
d
1 du u '
ln u 


dx
u dx u
u  x1
u’ = 1
1 1 
1
f '( x )  


2  x  1  2  x  1
Aim: Differentiating Natural Log Function
Course: Calculus
Model Problem
Differentiate f ( x )  ln
rewrite


x x 1
2

2
2 x3  1


1
 ln x  2ln x  1  2 x 3  1
2
2
d
1 du u ' u = x2 + 1
ln u 


u’ = 2x
dx
u dx u
1 
f '( x )  2 
x 
u = 2x3 – 1
u’ = 6x2
2

2x  1 6x 
  3

2
x  1  2  2 x  1 
1
4x
3 x2
  2
 3
x x  1 2x  1
Aim: Differentiating Natural Log Function

Course: Calculus
Logarithmic Differentiation
14
x of2 logs to simplify

Applying the laws
Differentiate y 
2
12
10
g x  =
 x-2 2


2 quotients, products
functions that include
x 1
and/or powers can simplify differentiation.
y is always positive
therefore ln y is defined
x 2+1
0.5
8
6
4
2
5
10
-2
x  2

ln y  ln
2
take ln of
both sides
x2  1

1
2
ln y  2ln  x  2   ln x  1
2
y'
 1  1  2x 
 2
  2


y
x

2
2
x

1




Aim: Differentiating Natural Log Function

Log properties
Differentiate
Course: Calculus
15
Using Log Derivative
x  2

Differentiate y 
y'
 1  1
 2
 

y
 x  2 2
2
x2  1
2
x
2x 

 2

2
x2 x 1
x 1
x 
 2
y'  y
 2

 x  2 x 1
Solve for y’,
 x2  2 x  2

2
2
x

2
x
 1
x  1  
2
x

2
x

  2x  2

32
2
x 1
x  2


2





Aim: Differentiating Natural Log Function


 Substitute for y

& Simplify
Course: Calculus
Derivative Involving Absolute Value
If u is a differentiable function of x
such that u  0, then
d
u'
 ln u  
dx
u
Find the derivative of f(x) = ln|cosx|
u = cosx
u’ = -sinx
d
u'
 ln cos x  
dx
u
 sin x

cos x
  tan x
Aim: Differentiating Natural Log Function
Course: Calculus
Model Problem

Find the relative extrema of y  ln x 2  2 x  3
u = x2 + 2x + 3
u'
2x  2
y' 
 2
u’ = 2x + 2
u x  2x  3
2x  2
1st Derivative
 0  x  1
2
Test
x  2x  3
f(-1) =
ln[-12
Evaluate
+ 2(-1) + 3] = ln 2 critical point
Relative Extrema
Minimum– (-1, ln2)
2 x 2  4 x  2
y '' 
2
2
x  2x  3



at x  1  positive result
concave up
Aim: Differentiating Natural Log Function
2nd Derivative
Test
6
g x  = lnx 2+2x+3
4
2
-5
Course: Calculus
5
Aim: How do we differentiate the natural
logarithmic function?
Do Now:
ln t
Find the derivative for g( t )  2
t
Aim: Differentiating Natural Log Function
Course: Calculus