Chabot Mathematics
§2.2 Methods of
Differentiation
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Review §
2.1
 Any QUESTIONS About
• §2.1 → Intro to Derivatives
 Any QUESTIONS About
HomeWork
• §2.1 → HW-07
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BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
§2.2 Learning Goals
 Use the constant multiple rule, sum
rule, and power rule to find derivatives
 Find relative and percentage rates of
change
 Study rectilinear motion and the
motion of a projectile
http://kmoddl.library.
cornell.edu/resource
s.php?id=1805
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Rule Roster
 Constant Rule
• For Any Constant c
d
c  0
dx
• The Derivative of any
Constant is ZERO
• Prove Using
Derivative Definition
df
f x  h   f x 
 lim
dx h0
h
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 For f(x) = c
dc
cc
 lim
h 0
dx
h
0
 lim  0
h 0 h
 Example  f(x) =73
• By Constant Rule
d
73  0
dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Rule Roster
 Power Rule
 
d n
n 1
x  nx
dx
• For any constant
real number, n
• Proof by Definition is VERY tedious, So Do
a TEST Case instead
• Let F(x) = x5; then plug into Deriv-Def
– The F(x+h) & F(x)
&
– Then: F(x+h) − F(x)
4
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3
2
1
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BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Rule Roster  Power Rule
• Then the Limit for h→0

0
0
0
0

dF
4
3
2 2
3
4
 lim h  5h x  10h x  10hx  5 x
dx h0
d 5
• Finally for d
n
4
n 1
x 
x  5 x  nx
n=5
dx
dx
 
 
• The Power Rule WILL WORK for every
other possible Test Case
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
MuPAD Code
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Rule Roster
lim k  f  x   k  lim f  x 
 Constant Multiple
Rule
x c
• Thus for the
Constant Multiplier
• For Any Constant c,
and Differentiable
Function f(x)
d
d
c  f x   c   f x 
dx
dx
• Proof: Recall from
Limit Discussion the
Constant Multiplier
Property:
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x c
d
 f x  h   f x  
c  f x   lim
c 

h

0
dx
h


 f x  h   f x  
 c  lim 

h 0
h


d
c  f x   c  d  f x 
dx
dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Rule Roster
 Sum Rule
• If f(x) and g(x) are Differentiable, then the
Derivative of the sum of these functions:
d
d
d
 f x   g x    f x   g x 
dx
dx
dx
• Proof: Recall from Limit Discussion the
“Sum of Limits” Property
lim  f  x   g x   lim f x   lim g x 
x c
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x c
x c
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Rule Roster  Sum Rule
• Then by Deriv-Def
d
 f  x  h   f  x  g  x  h g  x 
 f x   g x   lim 


dx
h
h


h 0
 f x  h   f x 
 g  x  h g  x  
 lim 
 lim 


h 0
h 0
h
h




d
d
  f  x   g  x 
dx
dx
• thus
d
d
d
 f x   g x    f x   g x 
dx
dx
dx
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BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Derivative Rules Summarized
d
c  0
If c is a constant,
dx
d n
n 1

If n is a constant,
x  n  x
dx
d
d
c  f ( x)  c   f ( x)
If c is a constant,
dx
dx
d
d
d
 f ( x)  g ( x)   f ( x)  g ( x)
dx
dx
dx
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BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Derivative Rules Summarized
 In other words…
• The derivative of a constant function is zero
• The derivative of a
constant times a
function is that
constant times the
derivative of
the function
• The derivative of the sum or difference of
two functions is the sum or difference of the
derivative of each function
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Derivative Rules: Quick Examples
 Constant Rule  d ( 7) = 0
dx
d 2
2-1
x ) = 2 × x = 2x
 Power Rule 
(
dx
 Constant
Multiple
Rule 
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d
2 2/3-1
2/3
-1/3
6x
=
6
×
x
=
4x
(
)
dx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  Sum/Diff & Pwr Rule
 Find df/dx
f x   x  2 x
for:
 SOLUTION
 Use the Difference & Power Rules
d
f '(x) =
dx
d
=
dx
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(
x - 2x
( )
)
d
x - ( 2x )
dx
(difference rule)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  Sum/Diff & Pwr Rule
 
 
d 0.5
d 1
f ' x  
x  2
x
dx
dx
= 0.5x
0.5-1
- 2x
0.5
2
 0.5 x
1-1
(constant multiple
rule)
(power rule)
 Thus
d
dx

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
x  2 x  0.5 x
 0.5
2x
2
2
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
RectiLinear (StraightLine) Motion
 If the position of an Object moving in a
Straight Line is described by the
function s(t) then:
• The Object
ds
VELOCITY, v(t) v t   s ' t  
dt
• The Object
ACCELERATION,
a(t)
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dv
a t   v' t  
dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
RectiLinear (StraightLine) Motion
 Note that:
• The Velocity (or Speed) of the Object is the
Rate-of-Change of the Object Position
• The Acceleration of the Object is the Rateof-Change of the Object Velocity
 To Learn MUCH MORE
about Rectilinear Motion
take Chabot’s PHYS4A
Course (it’s very cool)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
RectMotion: Positive/Negative
• Negative v → object is
moving to the LEFT
 For the Position
Fcn, s(t)
• Positive v → object is
moving to the RIGHT
• Negative s → object
is to LEFT of Zero
Position
• Positive s → object
is to RIGHT of Zero
Position
 For the Acceleration
Fcn, a(t)
• Negative a → object is
SLOWING Down
• Positive a → object is
SPEEDING Up
 For the Velocity Fcn,
v(t)
 st 
-10
-9
-8
-7
-6
-5
-4
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 st 
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  High Diver
 A High-Diver’s height, in meters, above
the surface of a pool t seconds after
jumping is given by by Math Model
h(t )  4.9t  4.5t  10
2
 For this situation Determine
how quickly diver is rising
(or falling) after 0.2
seconds? After 1 second?
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  High Diver
 SOLUTION
 Assuming that the Diver Falls Straight
Down, this is then a Rect-Mtn Problem
• In other Words this a
Free-Fall Problem
 Use all of the Derivative
rules Discussed previously
to Calculate the derivative
of the height function
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  High Diver
 Using Derivative Rules
d
2
h' t  
 4.9t  4.5t  10  vt 
dt


d 2
d
d
= -4.9 × ( t ) + 4.5× ( t ) + (10)
dt
dt
dt
 4.9  2t  4.5 1  0
 Thus
vt   h' t   9.8t  4.5 in units of m/s 
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  High Diver
 Use the Derivative fcn for v(t) to find
v(0.2s) & v(1s)
v0.2  h' 0.2  9.8(0.2)  4.5  2.54 m/s
• The POSITIVE velocity indicates that the
diver jumps UP at the Dive Start
v1  h' 1  9.81  4.5  5.3 m/s
• The NEGATIVE velocity indicates that the
diver is now FALLING toward the Water
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Relative & %-age Rate of Change
 The Relative Rate of
dQ dz
Change of a Quantity RoC rel 
Q z 
Q(z) with Respect to z:
 The Percentage RoC is simply the
Relative Rate of Change Converted to
the PerCent Form
• Recall that 100% of SomeThing is 1 of
SomeThing
100% dQ dz 100%
RoC %  RoC rel 


1
Q z 
1
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Relative RoC, a.k.a. Sensitivity
 Another Name for the Relative Rate of
Change is “Sensitivity”
 Sensitivity is a metric that measures
how much a dependent Quantity
changes with some change in an
InDependent Quantity
relative to the
BaseLine-Value of
the dependent Quantity
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
MultiVariable Sensitivty Analysis
B. Mayer, C. C. Collins, M. Walton,
“Transient Analysis of Carrier Gas
Saturation in Liquid Source Vapor
Generators”, Journal of Vacuum
Science Technolgy A, vol. 19,
no.1, pp. 329-344, Jan/Feb 2001
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Sensitivity: Additional Reading
 For More Info on Sensitivity see
• B. Mayer, “Small Signal Analysis of Source
Vapor Control Requirements for APCVD”,
IEEE Transactions on Semiconductor
Manufacturing, vol. 9, no. 3, pp. 344-365,
1996
• M. Refai, G. Aral, V. Kudriavtsev, B. Mayer,
“Thermal Modeling for APCVD Furnace
Calibration Using MATRIXx“,
Electrochemical Soc. Proc., vol. 97-9, pp.
308-316, 1997
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  Rice Sensitivity
 The demand for rice
in the USA in 2009
approximately
followed the function
100
D(p) =
,
p
 Use this Function to
find the percentage
rate of change in
demand for rice in
the United States at
a price of 500
dollars per ton
• Where
– p ≡ Rice Price in
$/Ton
– D ≡ Rice Demand in
MegaTons
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
Example  Rice Sensitivity
 SOLUTION
 Using Derivative Rules
 By %-RoC Definition
dD dp
RoC %  100
D p 
 Calculate RoC at
p = 500

dD dp 500
RoC % 500   100
D500 
dD d 100 



dp dp  p 
dé
-1/2 ù
= ë100 p û
dp
=100 × - 2 p
1
 50 p
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3 / 2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
-3/2
Example  Rice Sensitivity
 Finally evaluate the percentage rate of
change in the expression at p=500:

dD dp 500
100
D(500)
-3/2
-50(500)
= 100


0
.
1
%
100
500
 In other words, at a price of 500 dollars
per ton demand DROPS by 0.1% per
unit increase (+$1/ton) in price.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
WhiteBoard Work
 Problems From §2.2
• P60 → Rapid
Transit
• P68 → Physical
Chemistry
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BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
All Done for Today
Power
Rule
Proof
A LOT of Missing Steps…
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BMayer@ChabotCollege.edu • MTH15_Lec-07_sec_2-2_Differeniatation-Methods_.pptx
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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