Chabot Mathematics
§4.3 Exp & Log
Derivatives
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Review §
4.2
 Any QUESTIONS About
• §4.2 → Logarithmic Functions
 Any
QUESTIONS
About
HomeWork
• §4.2 → HW-19
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
§4.3 Learning Goals
 Differentiate exponential and logarithmic
functions
 Examine applications involving
exponential and logarithmic derivatives
 Employ logarithmic differentiation
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Derivative of ex
 For any Real
Number, x
 
d x
x
e e
dx
 Thus the ex fcn has the unusual
property that the derivative of the fcn is
the ORIGINAL fcn
• The proof of this is quite complicated. For
our purposes we treat this as a formula
– For a good proof (in Appendix) see:
 D. F. Riddle, Calculus and Analytical Geometry,
Belmont, CA, Wadsworth, 1974, pp. 325-331
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Derivative of ex
 Using the “repeating” nature of d(ex)/dx
d x
d d x 
x
x
x
y  e  y' 
e  e  y' ' 
e e

dx
dx  dx

 Meaning of Above: for any x-value, say
x = 1.9, All of these y-related quantities
are equal at e1.9 = 6.686
 
 
• The y CoOrd: 1.9, 6.686
• The Slope: dy dxx 1.9  6.686

• The ConCavity: d 2 y dx 2
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
x 1.9
 6.686
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  ex Derivative
 Differentiate: y  e x 1  x
 Using Rules d
d x
y
e 1 x
• Product
dx
dx
• Power
dé
d é xù
x
ù
= e × ë 1- x û + ëe û 1- x
x
• e
dx
dx
x 1
-1/2
x
= e × (1- x) × (-1) + e 1- x
2

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
dy
1 x
1 / 2
x
  e (1  x)  e 1  x
dx
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Chain Rule for eu(x)
 If u(x) is a differentiable function of x
then
d ux
d
du
ux


e 
e 
dx
du x 
x 
dx
 Using the ex derivative property
d
du x  u  x  du x 
ux
e

e 
du x 
dx
dx
d u
u du
u

Or
e  e
 e  u ' x 
dx
dx
 
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Tangent Line
 Find the equation of the
tangent line at x = 0
f x
for the function:
 SOLUTION:
 Use the Point-Slope Line Eqn,
y-yAP = m(x-xAP), with
  e

• Anchor Point, (xAP,yAP): 0, f 0  0, e
x2  x
0 2 0
  0, 1
• Slope at the Anchor Point: m0   df dx x 0
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Tangent Line
 Find Slope
d x -x
m=
e
at x = 0
dx
 Let: u  x 2  x
 Then: e x  x  eu
( )
2
x=0
2
 
d u
 Thus: e  eu
du


d
d 2
u
x  x  2x 1
dx
dx
and
 And by
d x  x d eu  du


Chain Rule m  dx e
du dx
 
2
m  e  2 x  1  e
u
Chabot College Mathematics
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x2  x
 2 x  1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Tangent Line
 Then m at x = 0
d x2  x
0 2 0
m
e
 e  2  0  1  1 1  1
dx
x 0
 
 Using m and the y  y  mx  x 
AP
AP
Anchor-Point in
y -1= -1( x - 0)
the Pt-Slope Eqn
 Convert Line-Eqn
y


x

1
to Slope-Intercept form
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Tangent Line
 Tangent Line at (0,1) Graphically
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Derivative of ln(x) = loge(x)
1
 For any POSITIVE d


ln
x

Real Number, x
dx
x
 Thus the ln(x) fcn has the unusual
property that derivative Does NOT
produce another Log
• The proof of this is quite complicated. For
our purposes we treat this as a formula
– For a good proof (in Appendix) see:
 D. F. Riddle, Calculus and Analytical Geometry,
Belmont, CA, Wadsworth, 1974, pp. 325-331
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  ln Derivative
 Find the Derivative of: y  f x   x
 Using Rules
2

dy d
x 
• Quotient
 

dx dx 1  ln x 
• Power
• ln(x)
=
2
1  ln x 
( x)
(1+ ln x)× 2x - x 2 1
(1+ ln x )
2
x  2 x ln x

2
1  ln x 
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Chain Rule for ln(u(x))
 If u(x)> 0 is a differentiable function of x
then d
d
du x 
dx
ln ux  
du x 
ln ux 
 Using the ln(x) derivative property
dx
d
du x 
1 du x 
ln ux 


du x 
dx
u x  dx
d
1 du
1
ln u  
Or

 u ' x 
dx
u dx
u
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Derivative of ax & loga(x)
 For Base a with
d x
x
a>0 and a≠1,
a  ln a a
then for ALL x: dx
 For Base a with
d
1
log a x  
a>0 and a≠1, then
dx
x  ln a
for ALL x>0:
 Prove Both on White/Black Board
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Revenue RoC
 The total number of hits (in thousands)
to a website t months after the
beginning of 1996 is modeled by
H t   200 ln t  4
 The Model for the weekly advertising
revenue in ¢ per hit: r t   25  0.1t
 Use the Math Models to determine the
daily revenue change at the beginning
of the year 2005
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Revenue RoC
 SOLUTION:
 The rate of change in Total Revenue,
R(t), is the Derivative of the Product of
revenue per hit and total hits:
d
d
d
Rt   H t   r t   200 ln( t  4)  (25  0.1t )
dt
dt
dt
d
d
= [ 200 ln(t + 4)] × (25 - 0.1t)+ [(25- 0.1t)] × 200 ln(t + 4)
dt
dt
1
= 200
×1× (25- 0.1t) + -0.1× 200 ln(t + 4)
t+4
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Revenue RoC
 Thus dRt   R' t   5000  20t  20 ln( t  4)
dt
t4
 Next find t in months for 1996→2005
12 mon
t  2005  1996 yrs 
 108 mon
1 yr
 Then the rate derivative at t = 108 mon
dR
5000  20(108)

 20 ln( 108  4) » -69.0128.
dt 108
108  4
 A units dR R H  r  kHits  Cent



t
mon
analysis dt t
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Hit

kCent
mon
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Revenue RoC
 The units on H are kHits, and units on r
are ¢/Hit. The units on time were
months so the derivative has units
k¢/mon. Convert to $/mon:
kCent 10 $
$
 69.013

 690.13
mon kCent
mon
 STATE: at the beginning of 2005 the
website was making about $690.13
LESS each month that passed.
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Helpful Hint  Log Diff
 Logarithmic Differentiation
 Some derivatives are easier to calculate
by
• first take the natural logarithm of the
expression
• Next judiciously use the log rules
• then take the derivative of both sides of the
equation
• finally solve for the derivative term
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Using Log Diff
 Using logarithmic
3 x2 3
2
differentiation to
f ( x)  3x e  x  1
find the df/dx for:
 SOLUTION:
 Computing the derivative directly would
involve the repeated use of the product
rule (not impossible, but very tedious)
 Instead, use properties of logarithms to
first expand the expression
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Using Log Diff
 Let y = f(x) → f x   y  3x3e x2  3 x 2  1
 Then take the natural logarithm of both
sides:
ln y = ln é3x 3e x-2 × 3 x 2 +1ù
ë
û
3
x-2
3 2
ln y = ln3+ ln x + lne + ln x +1
 Use the Power
1
2
ln
y
=
ln3+
3ln
x
+
x
2
+
ln
x
+1)
(
& Log Rules
3
 Now Take the Derivative of Both Sides
ù
d
dé
1
2
ln y = êln3+ 3ln x + x - 2 + ln ( x +1)ú
û
dx
dx ë
3
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Using Log Diff
 By the Chain Rule
 d dy 
d
d
dy 1 dy
ln y    ln y   ln y    
dx
dy
dx y dx
 dy dx 
 Then
d
1 dy d 
1

2
ln y    ln 3  3 ln x  x  2  ln x  1
dx
y dx dx 
3

1 dy 3
1 1
 Or    1   2  2 x
y dx x
3 x 1
• This is a form of Implicit Differentiation;
Need to algebraically Isolate dy/dx
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Example  Using Log Diff
 Solving 1  dy  3  1  1  1  2 x
2


y
dx
x
3
x
1
for dy/dx
dy
2x 
3
  y    1  2

dx
3x  3 
x
3 x2 3
 Recall y  3x e
 x  1 and
2

dy df x 
y  f x  

dx
dx

df x 
2x 
3
3 x2 3
2
 3x e  x  1    1  2

 Thus f ' x  
dx
3x  3 
x
• This result would have much more
difficult to obtain without the use of the
Log transform and implicit differentiation
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
WhiteBoard Work
 Problems From §4.3
• P76 → Per Capita Growth
• P90 → Newtons Law of (convective)
Cooling
– Requires a Biot Number* of Less than 0.1
INternal Thermal Resistance
Bi →
EXternal Thermal Resistance
*B. V. Karlekar, R. M.
Desmond, Engineering
Heat Transfer, St. Paul,
MN, West Publishing Co.,
1977, pp. 103-110
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
All Done for Today
For
PHYS4A
Students
From RigidBody Motion-Mechanics
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
ConCavity Sign Chart
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
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−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
x
Summary of Log Rules
 For any positive numbers M, N, and a
with a ≠ 1, and whole number p
log a ( MN )  log a M  log a N ;
log a M
p
 p log a M ;
Power Rule
M
log a
 log a M  log a N ;
N
k
log a a  k .
Chabot College Mathematics
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Product Rule
Quotient Rule
Base-to-Power Rule
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Change of Base Rule
 Let a, b, and c be positive real
numbers with a ≠ 1 and b ≠ 1.
Then logbx can be converted to a
different base as follows:
log a x
log x
ln x
log b x 


log a b
log b
ln b
(base a) (base 10) (base e)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Derive Change of Base Rule
 Any number >1 can be used for b, but since
most calculators have ln and log functions we
usually change between base-e and base-10
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Prove d(ex)/dx =ex
– D. F. Riddle, Calculus and Analytical Geometry,
Belmont, CA, Wadsworth, 1974, pp. 325-331
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Prove d(ex)/dx =ex
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BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
 D. F. Riddle, Calculus
and Analytical
Geometry, Belmont,
CA, Wadsworth,
1974, pp. 325-331
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-20_sec_4-3_EXP-n-LOG_Derivatives.pptx
Prove:
d x
a  ln a a x
dx
Chabot College Mathematics
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d
1
log a x  
dx
x  ln a
Bruce Mayer, PE
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