Transcript 11.2: Derivatives of exponential and Logarithmic Functions

11.2: Derivatives of Exponential
and Logarithmic Functions
Use the limit definition to find the derivative of ex
Find
f ( x  h)  f ( x )
h


f ( x  h)  f ( x ) e x  h  e x e x e h  e x e x e h  1



h
h
h
h
Find
f ( x  h)  f ( x )
lim
h 0
h
e (e  1)
e 1 x
x
 lim
 e lim
e
h 0
h 0
h
h
x
h
Because
h
eh 1
lim
1
h 0
h
Use graphing calculator
The Derivative of
x
e
Therefore: The derivative of f (x) = ex is f ’(x) = ex.
Example 1
Find f’(x)
A) f(x) = 4ex – 8x2 + 7x - 14
f’(x) = 4ex – 16x + 7
B) f(x) = x7 – x5 + e3 – x + ex
f’(x) = 7x6 – 5x4 + 0 –x + ex
= 7x6 – 5x4
–x + ex
Example 2
Find derivatives for
A) f (x) = ex / 2
f ’(x) = ex / 2
B) f (x) = 2ex +x2
f ’(x) = 2ex + 2x
C) f (x) = -7xe – 2ex + e2
f ’(x) = -7exe-1 – 2ex
Remember that e is a real
number, so the power
rule is used to find the
derivative of xe.
Also e2  7.389 is a
constant, so its derivative
is 0.
Review
y  logb x
is equivalent to
xb
y
Domain: (0, ∞)
Range: (0, ∞)
Range: (-∞, ∞)
Domain: (-∞, ∞)
* These are inverse function. The
graphs are symmetric with respect
to the line y=x
x  by
y  logb x
* There are many different bases for
a logarithmic functions. Two special
logarithmic functions are common
logarithm (log10x or log x) and
natural logarithm (logex = ln x)
Review: properties of ln
1)
2)
3)
4)
5)
ln(ab)  ln a  ln b
a
ln  ln a  ln b
b
k
ln a  k ln a
ln e  1
ln 1  0
Optional slide:
Use the limit definition to find the derivative of ln x
Find
f ( x  h)  f ( x )
ln(x  h)  ln x 1  x  h  Property 2

  ln

h
h
h
x 
x 1 xh
   ln
Multiply by 1 which is x / x

x h
x 
1  x  h 
   ln1   
x  h  x 
1
x


1   h  h   1   ln1  s  s
   ln1   
x   x   x 


Set s = h / x
Property 3
So when h approaches 0, s
also approaches o
s 
f ( x  h)  f ( x )
1 
 lim   ln1  s  
lim
s 0 x
h 0
h


1
1
s
1 

s 
1
 lim ln1  s    ln lim1  s 
s 0
x s 0 
 x 
1




 1
  ln e
 x

Definition of e
1

x
Property 4: ln(e)=1
The Derivative of ln x
Therefore: The derivative of f (x) = ln x is f ’(x) =
1
x
Example 3
Find y’ for
A) y  10x3 100ln x
100
1
2
2
y'  30x  100   30x 
x
 x
B) y  ln x5  e x  ln e2
y  5 ln x  e x  ln e2
1 x
y '  5   e  0
 x
5
  ex
x
More formulas
The derivative of f(x) = bx
is f’(x) = bx ln b
The derivative of f(x) = logb x
1 1 
is f’(x) = x  ln b 


Proofs are on page 598
Example 4
Find g’(x) for
A) g ( x)  x10  10x
g ' ( x)  10x9  10x ln(10)
B)
g ( x)  log2 x  6 log5 x
1  1   1  1 
g ' ( x)  
  6 

x  ln 2   x  ln 5 
1 1
6 
g ' ( x)  


x  ln 2 ln 5 
Example 5
An Internet store sells blankets. If the price-demand
equation is p = 200(0.998)x, find the rate of change of price
with respect to demand when the demand is 400 blankets
and explain the result.
p’ = 200 (.998)x ln(0.998)
p’(400) = 200 (.998)400 ln(0.998) = -0.18.
When the demand is 400 blankets, the price is decreasing
about 18 cents per blanket
Example 6
A model for newspaper circulation is C(t) = 83 – 9 ln t
where C is newspaper circulation (in millions) and t is the number of
years (t=0 corresponds to 1980). Estimate the circulation and find the
rate of change of circulation in 2010 and explain the result.
t = 30 corresponds to 2010
C(30) = 83 – 9 ln30 = 52.4
9
1
C(t)’ =  9   
t
t 
C’(30) =
9

 0 . 3
30
The circulation in 2010 is about 52.4 million and is
decreasing at the rate of 0.3 million per year
Example 7:
Find the equation of the tangent line to the graph
of f = 2ex + 6x at x = 0
Y = mx + b
f’(x) = 2ex + 6
m = f’(0) = 2(1) + 6 = 8
y=f (0) = 2(1) + 6(0) = 2
Y = mx + b
2 = 8(0) + b so b = 2
The equation is y = 8x + 2
Example 8:
Use graphing calculator to find the points of
intersection
F(x) = (lnx)2 and g(x) = x
On your calculator, press Y=
Type in the 2 functions above for Y1 and Y2
Press ZOOM, 6:ZStandard
To have a better picture, go back to ZOOM, 2: Zoom In
*Now, to find the point of intersection (there is only 1 in this problem),
press 2ND, TRACE then 5: intersect
Play with the left and right arrow to find the linking dot, when you see it,
press ENTER, ENTER again, then move it to the intersection, press
ENTER. From there, you should see the point of intersection
(.49486641, .49486641)