HCMUT –DEPARTMEND OF MATH. APPLIED
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CALCULUS FOR BUSINESS –221 Semester
CHAPTER 3: DERIVATIVE
•PhD. NGUYỄN QUỐC LÂN (November 2022)
RATE OF CHANGE: CONSTANT & VARIABLE
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Linearfunctiony=ax+bchangesatconstantrate→Slope
Example:Sincebeginningof
theyear,thepriceofabottle
ofsodahasbeenrisingata
constantrateof2
cents/month Price
functiony=2x+b(x:
numberofmonth)…
Generally,therateofchangeisnotconstant.Howtodescribe?
Answer:Usederivativef’(x)(whichistheslopeoftangentline).
-
F
General
+(x)
=>
Rate
=
f(x)
=
S
-
-------
PRACTICAL EXAMPLE
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DERIVATIVE
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Derivativeoff(x),x=a:
Reading:f’(a)is“fprimeofa”,anddf/dxis“deefbydeex”.
=
(a)
-
! I
TECHNIQUE (HOFFMANN, CHAPTER 2, SECTION 2 →4).
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() () () ()()
/
//1
/
2
/
3
/
2
,,,...3,2:Highschool
===
−
v
u
uvvuxxxxxx
Tangentequationof(C):y=f(x)atx=a:
Rateatwhichy=f(x)changeswithrespecttoxatx=a:
()()()axafafy −=−
/
()af
/
()
()
()
12
2
/
0
2
1
2
0
BKEL Calculated questions
21
a/ Given ,evaluate approximately
1
b/ Let be the tangent equation of (C): 2 3 at 1.
Evaluate approximately .
x
y y x dx
x
ygx y x xx
g x dx
+
=
+
==+=
12x
+
3x)'
=
4x
+3(or)
=
vv
+
uv
(cost)
=
-
sinx
a)y
=
x
+
1
↑
j
=
/
y
=I
5
b)y
=
2
x
3+3xat
x=
1
y(1)
=
5
y
=
6x2
+
3y'(1)
=
9
y
-
5
=
9(x
-
1)
=>
y
=
gx
-
4
=
1!
S
19x
-
4) 2dx
=
=
=
x)
b)y
=
==
y'(2)
=
-
3
c)FindWate
(x
-
1)2 y
=
xXchanges
at
x
=
1
,
5
=
2
a)
y
=
f()
=
2
+
3x
=
f()
=
5
Bam may
:)k
=
2
*
Eme
(x
=
15
=
9,45
=>
f
' (1)
=
7
=>
Tangent
:
y
-
5
=
7(x
-
1)
=
y
=
7x
-
2
+(g(x)dx
·
may
RATE OF CHANGE
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21
Example: Find the rate at which changes with respect to at 2
1
x
yxx
x
+
==
−
Rateatwhichy=f(x)changeswithrespecttoxatx=a:()af
/
Ans
:
P'(x)
=
-
800x
+
6800
=
4'/9)
=
-
400
dollars/thounds
unit
Conclusion
:
Profitdecreasing
atrateof400
dollars/thousands
units
p'(g)
lim
Dollars
ds
-
>
unitoffunction
~
variable
-
changeofcost
when
x
=
a
1
unit
((x)
=
292
+
49
,
970
2)
Use
marginal
costtoestimatethecostof
15th
un
it
1)
Marginal
costat
4?
((is)
-
((14)
=
C'(14)
C'(q)
=
4q
+4
=
C(4)
=
20
=
cost
increase
by20$
T
I
-
15 14 14
When
a
increase
from4
by
T
unit Nhof(a
+
1)
-
f(a)
=
f'(a)
Output
a
=
f(x
,
y)
Q
=
2x
+
xy
+
y x
:
numbersofhoursof
skill
labors
y
:
numbersofhoursofunskill
labors
x
=
30
x
↓bythour
-
>
a
=
constant
=>
2x3
+
x
y
+
y
=
const
y
=
20
&
yt
by
how
much
&
um
y
=
y(x)
=
y
-y
=
y
-
0x
Xx
=
1
(2x3
+
y
+
y))
=
(const)
=
6x2
+
2xy
+
x
y3yy
=
0by
=
y
.x
I
=-
3
,
14
=>
y
=
eye-3 =>
Decreasey inabout
3
,
14hours
C
=
C(q)
unitF(c
,
q)
=
const
-
>
Implicit
C
=
Clq)
If
g
,
c
change/respect?[C
-
39
=
4275
Chain
a
9
,
9
=
g(t)
, d
I
give
n
=C=?
1500units
=>q
=
15=C
=27
=
120Cost
=
120000dollars
(c
2
-
39)
!
=
142751
++
2CC
-
99q
=
0
=>
2x120xc
-
9x15x
=>
c
=
1
,
6875
=>
Cost
increases
withrateof
1687
.
5
dollars
/week
.
RELATED RATES. EXAMPLE 2.6.8 (SECTION 2.6)
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RELATED RATES. EXAMPLE 2
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EXAMPLE
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end
answer
x
+
0
,
=
8
continuous
ea
=
flo)lint
ing(x)
=
1
=
g(0)
=
+(
=
g()vxref
6
First
,
a
=
f(0)
=
limf(x)
x-0e
! 1x
-
>
0
=
I=L
When
a
=
1
+
d
=
INCREASING –DECREASING FUNCTION (HOFFMANN, CH. 3)
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RELATIVE (LOCAL) EXTREMA & MIN -MAX
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(C):y=f(x)hasarelative(local)maximumatx=ciff(c)f(x)x
nearc;relative(local)mininumatx=ciff(c)f(x)xnearc.
Relativemax,minarecalledrelativeextremaandareonlylocal.
EXTREMA & MIN MAX EXAMPLE
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INFLECTION POINT OR DIMINISHING RETURN
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Anefficiencystudyofthemorningshiftbetween8:00A.M.and
12:00noonatafactoryindicatesthatanaverageworkerwillhave
producedQ(t)=–t
3
+9t
2
+12tthourslater.Atwhattimeduring
themorningistheworkerperformingmostandleastefficiently?
(PRICE ELASTICITY OF DEMAND)
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EOD EXAMPLE
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CLASSIFICATION AND REVENUE
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EXAMPLE
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