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Mathematics for Business(Finance)
Instructor: Prof. Ken Tsang
Room E409-R11
Email: [email protected]
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Chapter 2: Differentiation( 微分 ): Basic Concepts
In this Chapter, we will encounter some important concepts.
The Derivative （导数）
Product （乘法准则） and Quotient Rules （除式准则） , Higher-Order Derivatives （高次导数 ）
The Chain Rule （链锁法则）
Marginal Analysis （边际分析） , Implicit Differentiation （隐函数求导） .
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Section 2.1 The Derivative （导数）
Calculus is the mathematics of change, and the primary tool for studying change is a procedure called differentiation （微分） .
In this section, we will introduce this procedure and examine some of its uses, especially in computing rates of change.
Rate of changes, for example velocity, acceleration, the changes, for example velocity, acceleration, the rate of growth of a population, and many others, are rate of growth of a population, and many others, are described mathematically by described mathematically by derivativesderivatives..
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Example 1
If air resistance is neglected, an object dropped from a If air resistance is neglected, an object dropped from a great height will fall feet in great height will fall feet in tt seconds. What seconds. What is the object’s instantaneous velocity after is the object’s instantaneous velocity after t=2t=2 seconds? seconds?
216)( tts
hh
h
h
shsVave
1664)2(16)2(16
2)2(
)2()2(
time elapsed
traveled distance
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Solution:Solution:
Average rate of change of Average rate of change of s(t)s(t) over the time period over the time period [2,2+h][2,2+h] by the ratioby the ratio
Compute the instantaneous velocity by the limitCompute the instantaneous velocity by the limit 64)1664(limlim
00
hVV
have
hins
That is, after 2 seconds, the object is traveling at the rate of 64 feet per second.
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Rates of Change （变化率） How to determine How to determine instantaneous rate of changeinstantaneous rate of change or rate or rate
of change of change of f(x)f(x) at at x=cx=c ? ?
Find the average rate of change of Find the average rate of change of f(x)f(x) as as xx varies from varies from x=cx=c to to x=c+hx=c+h
h
cfhcf
chc
cfhcf
x
xfrateave
)()(
)(
)()()(
Compute the instantaneous rate of change of Compute the instantaneous rate of change of f(x)f(x) at at x=cx=c by by finding the limiting value of the average rate as finding the limiting value of the average rate as hh tends to tends to 00
h
cfhcfraterate
have
hins
)()(limlim
00
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Rates of Change (Linear function)
A linear function A linear function y(x)=mx+by(x)=mx+b changes at the constant rate changes at the constant rate mm
with respect to the independent variable with respect to the independent variable xx. That is the rate of . That is the rate of
change of change of y(x)y(x) is given by the slope or steepness of its graph is given by the slope or steepness of its graph
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12
x in change
y in change
change of rateSlope
xx
yy
x
y
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Rates of Change (non-Linear function)For the function that is For the function that is nonlinear, nonlinear, the rate of change the rate of change is not constant but varies is not constant but varies with with xx..In particular, the rate of In particular, the rate of change at change at x=cx=c is given by the is given by the steepness of the graph of steepness of the graph of f(x)f(x) at the point at the point (c,f(c)),(c,f(c)), which which can be measured by can be measured by the slope the slope of the tangent line of the tangent line to the to the graph at graph at pp. .
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The slope of secant lineThe secant lineThe secant line （割线）（割线） : A line that intersects the curve at : A line that intersects the curve at
thethe
point point xx and point and point x+hx+h The average rate of change can be interpreted geometrica-The average rate of change can be interpreted geometrica-lly as the slope of the secant line from the point lly as the slope of the secant line from the point (x,f(x))(x,f(x)) to to the point the point (x+h,f(x+h)).(x+h,f(x+h)).
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The Derivative ( 导数 )A Difference Quotient for the function f(x): The express
The Derivative of a FunctionThe Derivative of a Function: The derivative of the : The derivative of the function function f(x)f(x) with respect to with respect to xx is the function is the function f’(x) f’(x) given given by by
The process of computing the derivative is called The process of computing the derivative is called differentiation. differentiation. f(x)f(x) is differentiable at is differentiable at x=cx=c if if f’(x)f’(x) exists exists
h
xfhxfxf
h
)()(lim)(
0
h
xfhxf )()(
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Example 2
Find the derivative of the function 35162)( 2 xxxf
Solution:Solution:
The difference quotient for f(x) is
16241624
3516235)(16)(2)()(
2
22
hxh
hhxh
h
xxhxhx
h
xfhxf
Thus, the derivative of f(x) is the function
164)1624(lim)()(
lim)(00
xhxh
xfhxfxf
hh
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Slope as a DerivativeSlope as a Derivative: The slope of the tangent : The slope of the tangent line to the curve line to the curve y=f(x)y=f(x) at the point at the point (c,f(c))(c,f(c)) is is
Instantaneous Rate of Change as a DerivativeInstantaneous Rate of Change as a Derivative: : The rate of change of The rate of change of f(x) f(x) with respect to with respect to x x when when x=cx=c is given by is given by f’(c)f’(c)
)(tan cfm
Remarks: Since the slope of the tangent line at (a,f(a)) is f’(a), the equation of the tangent line is
))(()( axafafy The point-slope form
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Example 3
What is the equation of the tangent line to the curve What is the equation of the tangent line to the curve
at the point where at the point where x=4 x=4 ??xy Solution:Solution:
According to the definition of derivative, we have
xxhxxhxh
xhxxhx
h
xhx
h
xfhxfxf
hh
hh
2
11lim
)(
))((lim
lim)()(
lim)(
00
00
Thus, the slope of the tangent line to the curve at the point where x=4 is f’(4)=1/4 . So the point-slope form is 1
4
1)4(
4
12 xyxy
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Example 4
Find the equation of the tangent line to the function Find the equation of the tangent line to the function at at x=16x=16.. xxxf 84)(
Solution:Solution:
We know that the equation of a tangent line is given byWe know that the equation of a tangent line is given by
So, we will need the derivative of the functionSo, we will need the derivative of the function
Now we need to evaluate the function and the derivative.Now we need to evaluate the function and the derivative.
Thus, the point-slope form for a tangent line is: Thus, the point-slope form for a tangent line is: 163)16(332 xyxy
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Example 5
A manufacturer determines that when A manufacturer determines that when xx thousand units of a thousand units of a particular commodity are produced, the profit generated will be particular commodity are produced, the profit generated will be
120006800400)( 2 xxxp
dollars. At what rate is profit changing with respect to the level of dollars. At what rate is profit changing with respect to the level of production production xx when when 99 thousand units are produced? thousand units are produced?
Solution:Solution:
We find thatWe find that
68008006800800400
lim
)120006800400(12000)(6800)(400lim
)()(lim)(
2
0
22
0
0
xh
hhxh
h
xxhxhx
h
xphxpxp
h
h
h
15
Thus, when the level of production is x=9, the profit is changing Thus, when the level of production is x=9, the profit is changing at the rate of dollars per at the rate of dollars per
thousand units. thousand units. 4006800)9(800)9( p
Which means that the tangent line to the profit curve Which means that the tangent line to the profit curve y=py=p((xx) is ) is sloped downward at point sloped downward at point QQ where where x=9x=9. Therefore, the profit . Therefore, the profit curve must be falling at curve must be falling at QQ and profit must be decreasing when and profit must be decreasing when 99 thousand units are being produced thousand units are being produced.
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Significance of the Sign of the derivative f’(x): If the function f is differentiable at x=c, then
f is increasing at x=c if f’(c)>0 and
f is decreasing at x=c if f’(c)<0
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Alternative Derivative Notation
Given a function Given a function y=fy=f((xx) all of the following are equivalent) all of the following are equivalent and represent the derivative of and represent the derivative of ff((xx) with respect to ) with respect to xx..
If we want to evaluate the derivative at If we want to evaluate the derivative at x=ax=a all of the all of the following are equivalentfollowing are equivalent
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Differentiability （可微） and Continuity （连续）
Continuity of a Differentiable FunctionContinuity of a Differentiable Function: If the function : If the function f(x)f(x) is differentiable at is differentiable at x=cx=c, then it is also continuous at , then it is also continuous at x=cx=c. .
Notice that a continuous function may not be differentiable Notice that a continuous function may not be differentiable
Graphs of four functions that are not differentiable at Graphs of four functions that are not differentiable at x=0x=0. .
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The Constant Rule: The Constant Rule: For any constant c, we have For any constant c, we have
That is, the derivative of a constant is zeroThat is, the derivative of a constant is zero
Section 2.2 Techniques of Differentiation
0cdx
d
0cdx
d
Proof:Proof: Since f(x+h)=c for all x
0lim)()(
lim)(00
h
cc
h
xfhxfxf
hh
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The Power RuleThe Power Rule For any real number For any real number nn
In words, to find the derivative of In words, to find the derivative of xxnn, reduce the , reduce the exponent exponent nn of of xx by by 11 and multiply your new power and multiply your new powerof of xx by original exponent. by original exponent.
1][ nn nxxdx
d
For Example
2
1
2
1
23
2
1)()(
3)(
xxdx
dx
dx
d
xxdx
d
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The constant Multiple RuleThe constant Multiple Rule If If cc is a constant and is a constant and f(x)f(x) is differentiable then so is is differentiable then so is cfcf((xx) and ) and
That is, the derivative of a multiple is the multiple of the That is, the derivative of a multiple is the multiple of the derivativederivative
)()( xfdx
dcxcf
dx
d
For Example
3344 12)4(3)(3)3( xxxdx
dx
dx
d
2/32/32/1
2
7)
2
1(7)7()
7(
xxxdx
d
xdx
d
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The Sum RuleThe Sum Rule: If : If ff((xx) and ) and gg((xx) are differentiable then so is the sum ) are differentiable then so is the sum of of ss((xx))=f=f((xx))+g+g((xx) and ) and
That is, the derivative of a sum is the sum of the separate derivativeThat is, the derivative of a sum is the sum of the separate derivative
)]([)]([)]()([ xgdx
dxf
dx
dxgxf
dx
d
3322 20)(2)7()()7( xxdx
dx
dx
dx
dx
d
For Example
84
847575
2110
)7(3)5(2)(3)(2)32(
xx
xxxdx
dx
dx
dxx
dx
d
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It is estimated that It is estimated that xx months form now, the months form now, the population of a certain community will bepopulation of a certain community will be
Example 6
800020)( 2 xxxp
a.a. At what rate will the population be changing with At what rate will the population be changing with respect to time respect to time 1515 months from now? months from now?
b.b. By how much will the population actually change By how much will the population actually change during the during the 1616thth month? month?
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Solution:Solution:
a.a. The rate of change of the population with respect to time The rate of change of the population with respect to time is the derivative of the population function. That isis the derivative of the population function. That is
202)(change of Rate xxp
The rate of change of the population 15 months from now The rate of change of the population 15 months from now will be will be month per people 5020)15(2)15( p
b. The actual change in the population during the 16th month is people 5185258576)15()16( pp
NoteNote: Since the rate varies during the month, the actual : Since the rate varies during the month, the actual change in population during 16change in population during 16 thth month differs from the month differs from the monthly rate of the change at the beginning of the month. monthly rate of the change at the beginning of the month.
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The Derivative as a Rate of Change (Review) The Derivative as a Rate of Change (Review)
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Relative and Percentage Rates of ChangeRelative and Percentage Rates of Change: The relative : The relative rate of change of a quantity rate of change of a quantity Q(x)Q(x) with respect to with respect to xx is is given by the ratiogiven by the ratio
The corresponding percentage rate of change of Q(x) with respect to x is
Q
dxdQ
xQ
xQ
Q(x)
/
)(
)(
of change
of rate Relative
)(
)(100
)( of change of
rate Percentage
xQ
xQ
xQ
Relative and Percentage Rates of ChangeRelative and Percentage Rates of Change
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Example 7
The gross domestic product (GDP) of a certain country was The gross domestic product (GDP) of a certain country was
billion dollars billion dollars tt years after 1995. years after 1995.
a.a. At what rate was the GDP changing with respect to time in At what rate was the GDP changing with respect to time in 2005?2005?
b.b. At what percentage rate was the GDP changing with respect to At what percentage rate was the GDP changing with respect to time in 2005? time in 2005?
1065)( 2 tttN
Solution:Solution:
a.a. The rate of change of the GDP is the derivative The rate of change of the GDP is the derivative N’(t)=2t+5. N’(t)=2t+5. The rate of change in The rate of change in 20052005 was was N’(10)=2(10)+5=25N’(10)=2(10)+5=25 billion billion dollars per year.dollars per year.
b.b. The percentage rate of change of the GDP in 2005 was The percentage rate of change of the GDP in 2005 was
yearper %77.9256
25100
)10(
)10(100
N
N
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Example 8
Let , find all Let , find all xx where where p’(x)>0p’(x)>0, , p’(x)=0p’(x)=0 and and p’(x)<0.p’(x)<0.
51232)( 23 xxxxp
Solution:Solution:
)2)(1(6
1266
012)2(3)3(2)(2
2
xx
xx
xxxp
Based on the techniques of Based on the techniques of differentiation, we have differentiation, we have
So the solution is
p’(x)=0 at x=-1 and 2
p’(x)>0 at x<-1 and x>2
p’(x)<0 at -1<x<2
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Rectilinear MotionRectilinear Motion: : Motion of an object along a straight-Motion of an object along a straight-line in some way. If the positionline in some way. If the position at time at time tt of an object of an object moving along a straight line is given by moving along a straight line is given by ss((tt)),, then the object then the object hashas
andand
The object is advancing when The object is advancing when v(t)>0v(t)>0, retreating when , retreating when v(t)<0v(t)<0, and stationary when v(t)=0. It is accelerating when , and stationary when v(t)=0. It is accelerating when a(t)>0a(t)>0 and decelerating when and decelerating when a(t)<0a(t)<0
dt
dststv )()( velocity
dt
dvtvta )()(on accelerati
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Example 9
The position at time t of an object moving along a line is given byThe position at time t of an object moving along a line is given by
a.a. Find the velocity of the object and discuss its motion between Find the velocity of the object and discuss its motion between
times times t=0t=0 and and t=4t=4..
b.b. Find the total distance traveled by the object between times Find the total distance traveled by the object between times t=0t=0 and and t=4t=4..
c.c. Find the acceleration of the object and determine when the Find the acceleration of the object and determine when the object is accelerating and decelerating between times object is accelerating and decelerating between times t=0t=0 and and t=4t=4. .
596)( 23 tttts
31
Solution:Solution:
The motion of an object:
596)( 23 tttts
Interval Sign of v(t)
Description of Motion
0<t<1 +Advancing
from s(0)=5 to s(1)=9
1<t<3 -Retreating
from s(1)=9 to s(3)=5
3<t<4 +Advancing
from s(3)=5 to s(4)=9
a.a. The velocity is . The object will be The velocity is . The object will be stationary whenstationary when
9123)( 2 ttdt
dstv
0)3)(1(39123)( 2 tttttv
that is, at times that is, at times t=1t=1 and and t=3t=3. Otherwise, the object is either . Otherwise, the object is either advancing or retreating, as described in the following table. advancing or retreating, as described in the following table.
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b.b. The object travels from The object travels from s(0)=5s(0)=5 to to s(1)=9s(1)=9, then back to , then back to s(3)=5s(3)=5, and finally to , and finally to s(4)=9s(4)=9. Thus the total distance . Thus the total distance traveled by the object istraveled by the object is
12|59||95||59|433110
ttt
D
c.c. The acceleration of the object is The acceleration of the object is
)2(6126)( ttdt
dvta
The object will be accelerating The object will be accelerating [a(t)>0][a(t)>0] when when 2<t<42<t<4 and and decelerating decelerating [a(t)<0][a(t)<0] when when 0<t<2.0<t<2.
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The Motion of a ProjectileThe Motion of a Projectile: An important example of : An important example of rectilinear motion . rectilinear motion .
Suppose an object is projected (e.g. thrown, fired, or Suppose an object is projected (e.g. thrown, fired, or dropped) vertically in such a way that the only dropped) vertically in such a way that the only acceleration acting on the object is the constant acceleration acting on the object is the constant downward acceleration downward acceleration g g due to gravity (that means due to gravity (that means air air resistance is negligibleresistance is negligible). Thus, the height of the object ). Thus, the height of the object at time at time tt is given by the formula is given by the formula
002
2
1)( HtVgttH
and where 00 VH are the initial height and velocity of theare the initial height and velocity of the
object, respectively.object, respectively.
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Example 10
Suppose a person standing at the top of a building Suppose a person standing at the top of a building 112112 feet high feet high
throws a ball vertically upward with an initial velocity of throws a ball vertically upward with an initial velocity of 9696 ft/sec. ft/sec.
a. Find the ball’s height and velocity at time t.
b. When does the ball hit the ground and what is its impact velocity?
c. When is the velocity 0? What is the significance of this time?
d. How far does the ball travel during its flight?
35
Solution:Solution:
a.a. Since the height of the ball above Since the height of the ball above the ground at time the ground at time tt is feet. The is feet. The velocity at time velocity at time tt is is
,112 and 96 ,32 00 HVg1129616)( 2 tttH
ft/sec 9632)( tdt
dHtv
b.b. The ball hits the ground when The ball hits the ground when H=0. H=0. Solve the equation we find Solve the equation we find t=7t=7 and and t=-1(discard)t=-1(discard). The impact velocity is The impact velocity is v(7)=-128 ft/sec v(7)=-128 ft/sec (The negative sign means the direction of motion is different with the direction of initial velocity. )
c.c. Set Set v(t)=0,v(t)=0, solvesolve t=3 t=3. Thus, the ball is at its highest point when . Thus, the ball is at its highest point when t=3t=3 seconds. seconds.
d.d. The ball starts at The ball starts at H(0)=112H(0)=112 feet and rises to a maximum height feet and rises to a maximum height of of H(3)=256H(3)=256. So . So
The total distance travelled=(256-112)+256=400 feet The total distance travelled=(256-112)+256=400 feet
36
The derivative of a product of functions is The derivative of a product of functions is not not the product of the product of separate derivative!! Similarly, the derivative of a quotient of separate derivative!! Similarly, the derivative of a quotient of functions is not the quotient of separate derivative. functions is not the quotient of separate derivative.
Section 2.3 Product and Quotient Rules; Higher-Order
Derivative
Suppose we have two functionSuppose we have two function f(x)=x f(x)=x33 and and g(x)=xg(x)=x66
37
The Product RuleThe Product Rule （乘法准则）（乘法准则） If the two functions If the two functions f(x)f(x) and and
g(x)g(x) are differentiable at are differentiable at xx, then we have the derivative of the , then we have the derivative of the
product product PP((xx))=f=f((xx))gg((xx) is ) is
or equivalently, or equivalently,
)]([)()]([)()()( xfdx
dxgxg
dx
dxfxgxf
dx
d
fggffg )(
Example 11
Using the product rule toUsing the product rule to find the derivative of the functionfind the derivative of the function
Solution:Solution:
3
5
3
2
3
5
3
2
3
5
3
2
3 223
1
3
8
3
1022
3
2
3
4
)22()2(3
2)(
xxxxxx
xxxxxxy
)2( 23 2 xxxy
38
A manufacturer determines that A manufacturer determines that tt months after a new months after a new product is introduced to the market, hundred product is introduced to the market, hundred units can be produced and then sold at a price ofunits can be produced and then sold at a price of
dollars per unit .dollars per unit .
a.a. Express the revenue Express the revenue R(t)R(t) for this product as a function for this product as a function of time .of time .
b. At what rate is revenue changing with respect to time b. At what rate is revenue changing with respect to time after after 44 months? Is revenue increasing or decreasing at months? Is revenue increasing or decreasing at
this time? this time?
Example 12
tttx 3)( 2
302)( 2
3
ttp
39
Solution:Solution:
a. The revenue is given by )302)(3()()()( 2/32 ttttptxtR
hundred dollars.
b. The rate of change of revenue R(t) with respect to time is given by the derivative R’(t), which we find using the product rule:
]32)[302()2
3(2)3(
]3[)302(302)3()(
2/32/12
22/32/32
ttttt
ttdt
dtt
dt
dtttR
At time t=4, the revenue is changing at the rate R’(4)=-14 Thus, after 4 months, the revenue is changing at the rate of 14 hundred dollars per month. It is decreasing at that time since R’(4) is negative.
40
The Quotient RuleThe Quotient Rule （除式准则）（除式准则） If the two functions If the two functions ff((xx) and ) and gg((xx) are differentiable at x, then the derivative of the quotient ) are differentiable at x, then the derivative of the quotient QQ((xx))=f=f((xx))/g/g((xx) is given by ) is given by
or equivalently, or equivalently,
0)( if )(
)]([)()]([)(]
)(
)([
2
xg
xg
xgdxd
xfxfdxd
xg
xg
xf
dx
d
2
''')(
g
fggf
g
f
Example 13
Using the quotient rule toUsing the quotient rule to find the derivative of the functionfind the derivative of the function
Solution:Solution:
22 )2(
15
)2(
)1)(93()2(3)(
zz
zzzW
41
Example 14
A biologist models the effect of introducing a toxin to a bacterial colony by the function
4
1)(
2
tt
ttP
where P is the population of the colony (in millions) t hours after the toxin is introduced.
a. At what rate is the population changing when the toxin is introduced? Is the population increasing or decreasing at this time?
b. At what time does the population begin to decrease? By how much does the population increase before it begins to decline?
42
Solution:Solution:
a. The rate of change of the population with respect to time is given by the derivative, that is
22
2
22
2
22
22
)4(
32
)4(
)12)(1()1)(4(
)4(
]4[)1(]1[)4()(
tt
tt
tt
tttt
tt
ttdtd
ttdtd
tttP
That is the population is initially changing at the rate of P’(0)=0.1875 million bacteria per hour, and it is increasing since P’(0)>0.
b. Since the numerator of P’(t) can be factored as –(t-1)(t+3), the denominator and the factor t+3 are both positive for all t≥0, which means that for 0≤t<1 P(t) is increasing, and for t>1, P(t) is decreasing. Therefore, before the population begins to decline, it increases by P(1)-P(0)=1/3-1/4=1/12 million.
43
A Word of Advice: The quotient rule is somewhat cumbersome, so don’t use it unnecessarily.
Example 15
Differentiate the function . x
xx
xy
1
5
4
33
22
Solution:Solution:
Don’t use the quotient rule! Instead, rewrite the function as 12 1
5
4
3
1
3
2 xxxy
and then apply the power rule term by term to get
2323
23
1
3
1
3
4
3
1
3
4
)1(003
1)2(
3
2
xxxx
xxdx
dy
44
The Second Derivative （二次导数） In applications, it may be necessary to compute the rate of change In applications, it may be necessary to compute the rate of change
of a function that is itself a rate of change. For example, the of a function that is itself a rate of change. For example, the acceleration of a car is the rate of change with respect to time of acceleration of a car is the rate of change with respect to time of its velocity. its velocity.
The second derivative of a function is the derivative The second derivative of a function is the derivative of its derivative. If of its derivative. If y=f(x)y=f(x), the second derivative is , the second derivative is denoted by denoted by
The second derivative gives the rate of change of the The second derivative gives the rate of change of the rate of change of the original function. rate of change of the original function.
)(or 2
2
xfdx
yd
Note: The derivative Note: The derivative f’(x)f’(x) is called the first is called the first derivative. derivative.
45
Alternate NotationAlternate Notation: There is some alternate notation for higher : There is some alternate notation for higher order derivatives as well.order derivatives as well.
2
2
)( )(dx
fdxf
dx
dfxf
Note: Before computing the second derivative of a function, always Note: Before computing the second derivative of a function, always
take time to simplify the first derivative as much as possible.take time to simplify the first derivative as much as possible. Example 16
An efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 8:00 A.M. will have produced units t hours later.
a. Compute the worker’s rate of production at 11:00 A.M.
b. At what rate is the worker’s rate of production changing with respect to time at 11:00 A.M.?
ttttQ 246)( 23
46
Solution:Solution:
a. The worker’s rate of production is the first derivative 24123)()( 2 tttQtR
of the output Q(t). The rate of production at 11:00 A.M. (t=3) is hour per units 33)3()3( QR
b. The rate of change of the rate of production is the second derivative of the output function. At 11:00 A.M., this rate is
The minus sign indicates that the worker’s rate of production is decreasing; that is, the worker is slowing down. The rate of this decrease in efficiency at 11:00 A.M. is 6 units per hour per hour.
126)()( ttQtR
hour per hour per units 6)3()3( QR
47
Example 17
The acceleration a(t) of an object moving along a straight line is the derivative of the velocity v(t). Thus, the acceleration may be thought of as the second derivative of position; that is 2
2
)(dt
sdta
If the position of an object moving along a straight line is given by at time t, find its velocity and acceleration. 43)( 23 ttts
Solution:Solution:
The velocity of the object is
and its acceleration is
463)( 2 ttdt
dstv
66)(2
2
tdt
sd
dt
dvta
48
The The nnth Derivative: th Derivative: For any positive integer For any positive integer nn, the , the nnth derivative th derivative of a function is obtained from the function by differentiating of a function is obtained from the function by differentiating successively successively nn times. If the original function is times. If the original function is y=fy=f((xx) the ) the nnth th derivative is denoted byderivative is denoted by
)(or )( xfdx
yd nn
n
Example 18
Find the fifth derivative of the function Find the fifth derivative of the function y=1/xy=1/x
Solution:Solution:
665
5
5
554
4
4
443
3
3
332
2
2
221
120120)24(
2424)6(
66)2(
22)(
1)(
xxx
dx
d
dx
yd
xxx
dx
d
dx
yd
xxx
dx
d
dx
yd
xxx
dx
d
dx
yd
xxx
dx
d
dx
dy
49
Section 2.4 The Chain Rule 链锁法则
)unitper dollars( output respect towith
cost of change of rate
dq
dC
hour)per (units timerespect to with
output of change of rate
dt
dq
Suppose the total manufacturing cost at a certain factory is a Suppose the total manufacturing cost at a certain factory is a function of the number of units produced, which in turn is a function function of the number of units produced, which in turn is a function of the number of hours the factory has been operating. If of the number of hours the factory has been operating. If C, q, t,C, q, t, denote the cost, units produced and time respectively, then denote the cost, units produced and time respectively, then
The product of these two rates is the rate of change of cost withThe product of these two rates is the rate of change of cost withrespect to time that isrespect to time that is
hour)per (dollars dt
dq
dq
dC
dt
dC
50
The Chain Rule: If If y=f(u)y=f(u) is a differentiable function of is a differentiable function of u u and and u=g(x) u=g(x) is in turn a differentiable function of is in turn a differentiable function of xx, , then the composite function then the composite function y=f(g(x))y=f(g(x)) is a differentiable is a differentiable function of function of x x whose derivative is given by the product whose derivative is given by the product
or, equivalently, by or, equivalently, by
dx
du
du
dy
dx
dy
)())(( xgxgfdx
dy
NoteNote: One way to remember the chain rule is to pretend : One way to remember the chain rule is to pretend the derivative the derivative dy/dudy/du and and du/dxdu/dx are quotients and to are quotients and to “cancel” “cancel” dudu. .
51
Example 19
Find if Find if dx
dy 1)2(3)2( 2232 xxy
Solution:Solution:
We rewrite the function as , where . Thus,
13 23 uuy
22 xu
xdx
duuu
du
dy2 and 63 2
and according to the chain rule,
)2(6)2]()[2(3
2 )2)](2(6)2(3[
)2)(63(
2322
2222
2
xxxxx
xwithureplacexxx
xuudx
du
du
dy
dx
dy
52
Example 20
The cost of producing The cost of producing xx units of a particular commodity units of a particular commodity is is
dollars, and the production level dollars, and the production level t t hours into a particular hours into a particular production run is production run is
5343
1)( 2 xxxC
units 03.02.0)( 2 tttx
At what rate is cost changing with respect to time after 4 hours?
53
Solution:Solution:
We find that We find that
So according to the chain rule, we have So according to the chain rule, we have
03.04.0 and 43
2 t
dt
dxx
dx
dC
)03.04.0(43
2
tx
dt
dx
dx
dC
dt
dC
When t=4, the level of production is x(4)=3.32 units, and by substituting t=4 and x=3.32 into the formula for
, we get dt
dC1277.10]03.0)4(4.0[4)32.3(
3
2
4
tdt
dC
Thus, after 4 hours, cost is increasing at the rate of approximately $10.13 per hour.
54
Let’s look at the functionsLet’s look at the functions
then we can write the function as a composition.
differentiate a composition function using the Chain Rule. differentiate a composition function using the Chain Rule.
The derivative is thenThe derivative is then
55
In general, we differentiate the outside function In general, we differentiate the outside function leaving the inside function alone and multiple all leaving the inside function alone and multiple all of this by the derivative of the inside function of this by the derivative of the inside function
function inside
of derivative alonefunction
inside leavefunction outside
of derivative
)( ))(( )( xgxgfxF
Exercise
a.a. Based on the chain rule, find the derivative of Based on the chain rule, find the derivative of
with respect to with respect to xx
b.b. Find the derivative of Find the derivative of
32 )()( xxxf
11
1)(
2
xxf
56
The General Power RuleThe General Power Rule:: For any real number For any real number nn and and differentiable function differentiable function hh, we have , we have
)]([)]([)]([ 1 xhdx
dxhnxh
dx
d nn
Think of [Think of [h(x)h(x)]]nn as the composite functionas the composite function nn uxhgxh g where)]([)]([
By the chain rule, we have By the chain rule, we have
)]([)]([)()]([)]([)]([ 1 xhdx
dxhnxhxhgxhg
dx
dxh
dx
d nn
57
An environmental study of a certain suburban An environmental study of a certain suburban community suggests that the average daily level community suggests that the average daily level of carbon monoxide in the air will beof carbon monoxide in the air will be parts per million when the parts per million when the population is p thousand. It is estimated that t population is p thousand. It is estimated that t years from now, the population of the years from now, the population of the community will be thousand. community will be thousand. At what rate will the carbon monoxide level At what rate will the carbon monoxide level be changing with respect to time be changing with respect to time 33 years from years from now?now?
Example 21
175.0)( 2 ppc
21.01.3)( ttp
58
Solution:Solution:
The goal is to find dc/dt when t=3. Since
0.2
and
)175.0(2
1)]2(5.0[)175.0(
2
1 2
122
12
tdt
dp
ppppdp
dc
it follows from the chain rule that
175.0
1.0)2.0()175.0(
2
12
2
12
p
pttpp
dt
dp
dp
dc
dt
dc
When t=3, p(3)=4 and substituting t=3 and p=4 into the formula for dc/dt, we get
year per million per parts 24.05
2.1
17)4(5.0
)3)(4(1.02
dt
dc
59
For instance, suppose For instance, suppose C(x)C(x) is the total cost of producing is the total cost of producing xx units of units of a particular commodity. If units are currently being produced, a particular commodity. If units are currently being produced, then the derivative then the derivative
Section 2.5 Marginal Analysis and Approximations Using Increments
In economics, the use of the derivative to approximateIn economics, the use of the derivative to approximate
the change in a quantity that results from a the change in a quantity that results from a 11-unit -unit
increase in production is called increase in production is called marginal analysis marginal analysis 边际分析边际分析
0x
h
xChxCxC
h
)()(lim)( 00
00
is called the marginal cost of producing units. The limiting value that defines this derivative is approximately equal to the difference quotient of C(x) when h=1; that is
0x
)()1(1
)()1()( 00
000 xCxC
xCxCxC
60
Marginal CostMarginal Cost （边际成本）（边际成本） :: If If C C((xx) is the total cost of ) is the total cost of producing producing xx units of a commodity. Then the units of a commodity. Then the marginal marginal costcost of producing of producing units is the derivative , which units is the derivative , which approximates the additional cost incurred approximates the additional cost incurred when the level of production is increased by one unit, when the level of production is increased by one unit, from to from to +1+1, assuming >>1., assuming >>1.
0x )( 0xC)()1( 00 xCxC
0x 0x 0x
61
The marginal revenue is , it approximatesThe marginal revenue is , it approximates , the additional revenue generated , the additional revenue generated by producing one more unit.by producing one more unit.
)( 0xR)()1( 00 xRxR
)( 0xP)()1( 00 xPxP
The marginal profit is , it approximatesThe marginal profit is , it approximates , the additional profit obtained by , the additional profit obtained by producing one more unit, assuming >>1.producing one more unit, assuming >>1.
Marginal Revenue （边际收入） and Marginal Profit （边际利润） : SupposeSuppose R R((xx) is the revenue function generated ) is the revenue function generated when when xx units of a particular commodity are units of a particular commodity are produced, and produced, and PP((xx) is the corresponding profit ) is the corresponding profit function, when function, when x=xx=x00 units are being produced, then units are being produced, then
0x
62
Example 22
A manufacturer estimates that when A manufacturer estimates that when xx units of a particular units of a particular commodity are produced, the total cost will be commodity are produced, the total cost will be dollars, and furthermore, that all x units will be sold when the price is dollars, and furthermore, that all x units will be sold when the price is dollars per unit. dollars per unit.
9838
1)( 2 xxxC
)75(3
1)( xxp
a. Find the marginal cost and the marginal revenue.
b. Use marginal cost to estimate the cost of producing the ninth unit.
c. What is the actual cost of producing the ninth unit?
d. Use marginal revenue to estimate the revenue derived from the sale of the ninth unit.
e. What is the actual revenue derived from the sale of the ninth unit?
63
Solution:Solution:
a.a. The marginal cost is The marginal cost is C’(x)=(1/4)x+3. C’(x)=(1/4)x+3. The total revenue is The total revenue is
, , the marginal revenue is the marginal revenue is R’(x)=25-2x/3. R’(x)=25-2x/3.
b.b. The cost of producing the ninth unit is the change in cost as The cost of producing the ninth unit is the change in cost as xx increases from increases from 88 to to 99 and can be estimated by the marginal cost and can be estimated by the marginal cost C’(8)=8/4+3=$5.C’(8)=8/4+3=$5.
c.c. The actual cost of producing the ninth unit is The actual cost of producing the ninth unit is C(9)-C(8)=$5.13C(9)-C(8)=$5.13
which is reasonably well approximated by the marginal cost which is reasonably well approximated by the marginal cost C’(8)C’(8)
d.d. The revenue obtained from the sale of the ninth unit is The revenue obtained from the sale of the ninth unit is approximated by the marginal revenue approximated by the marginal revenue R’(8)=25-2(8)/3=$19.67 R’(8)=25-2(8)/3=$19.67
e.e. The actual revenue obtained from the sale of the ninth unit is The actual revenue obtained from the sale of the ninth unit is
R(9)-R(8)=$19.33. R(9)-R(8)=$19.33.
3/25)3/)75(()()( 2xxxxxxpxR
64
Marginal analysis is an important example of a general Incremental approximation procedure
65
Based on the fact that since Based on the fact that since
h
xfhxfxf
h
)()(lim)( 00
00
then for small then for small h, h, the derivative is approximately the derivative is approximately equal to the difference quotient equal to the difference quotient . We indicate . We indicate this approximation by writing this approximation by writing
)( 0xf
h
xfhxf )()( 00
hxfxfhxfh
xfhxfxf )()()( ly,equivalent or,
)()()( 000
000
Or, equivalently, if , thenOr, equivalently, if , then
Approximation by IncrementApproximation by Increment If If ff((xx) is differentiable ) is differentiable at x=xat x=x00 and x is a small change in x, then△ and x is a small change in x, then△
xxfxfxxf )()()( 000
)()( 00 xfxxff
xxff )( 0
66
Example 23
During a medical procedure, the size of a roughly tumor During a medical procedure, the size of a roughly tumor is estimated by measuring its diameter and using the is estimated by measuring its diameter and using the formula to compute its volume. If the diameter formula to compute its volume. If the diameter is measured as is measured as 2.52.5 cm with a maximum error of cm with a maximum error of 2%,2%, how how accurate is the volume measurement?accurate is the volume measurement?
3
3
4RV
Solution:Solution:
A sphere of radius A sphere of radius RR and diameter and diameter x=2Rx=2R has volume has volume
33333 cm 181.8)5.2(6
1
6
1)
2(
3
4
3
4 x
xRV
The error made in computing this volume using the diameter The error made in computing this volume using the diameter 2.52.5
when the actual diameter is when the actual diameter is 2.5+2.5+△△x x is is xVVxVV )5.2()5.2()5.2(
To be continued
67
))5.2(02.0)](5.2([
)]5.2([Vin volumeerror Maximum
V
xV
The corresponding maximum error in the calculation of The corresponding maximum error in the calculation of volume is volume is
Since Since
817.9)5.2(2
1)5.2(
2
1)3(
6
1)( 222 VxxxV
it follows that it follows that 491.0)05.0)(817.9(in volumeerror Maximum
Thus, at worst, the calculation of the volume as Thus, at worst, the calculation of the volume as 8.1818.181
is off by is off by 0.4910.491 , so the actual volume , so the actual volume VV must satisfy must satisfy
3cm
3cm
672.8690.7 V
68
The percentage change of a quantity expresses the changeThe percentage change of a quantity expresses the change in that quantity as a percentage of its size prior to the in that quantity as a percentage of its size prior to the change. In particular,change. In particular,
quantity of size
quantityin change100change of Percentage
69
Example 24
The GDP of a certain country was billion dollars The GDP of a certain country was billion dollars tt years after years after 19971997. Use calculus to estimate the percentage change in . Use calculus to estimate the percentage change in
the GDP during the first quarter of the GDP during the first quarter of 20052005..
2005)( 2 tttN
Solution:Solution:
Use the formula Use the formula
)(
)(100in change Percentage
tN
ttNN
with with t=8, t=8, △△t=0.25t=0.25 and and N’(t)=2t+5N’(t)=2t+5, to get , to get
%73.1200)8(5)8(
)25.0](5)8(2[100
)8(
25.0)8(100in change Percentage
2
N
NN
70
DifferentialsDifferentials （微分）（微分） : The differential of : The differential of xx is is dx=dx=△△xx, , and if and if y=f(x)y=f(x) is a differentiable function of is a differentiable function of xx, then , then dy=f’(x)dxdy=f’(x)dx is the differential of is the differential of yy. .
71
Example 25
In each case, find the differential of In each case, find the differential of y=f(x)y=f(x). .
27)( 23 xxxfa.a.
b.b. )23)(5()( 22 xxxxf
Solution:Solution:
a. a.
b. By the product rule,b. By the product rule,
dxxxdxxxdxxfdy )143()]2(73[)( 22
dxxxx
dxxxxxxdxxfdy
)51438(
)]23)(2()41)(5[()(23
22
72
Section 2.6 Implicit Differentiation （隐函数求导） and Related Rates
23
2 1 and 32
1 13 xy
x
xyxxy
So far the functions have all been given by equations of the form So far the functions have all been given by equations of the form y=f(x)y=f(x). A function in this form is said to be in . A function in this form is said to be in explicit formexplicit form. .
For example, the functions For example, the functions
are all functions in are all functions in explicit formexplicit form （显式表达式）（显式表达式） ..
Sometimes practical problems will lead to equations in which Sometimes practical problems will lead to equations in which the function the function yy is not written explicitly in terms of the independent is not written explicitly in terms of the independent variable variable x. x. For example, the equations such as For example, the equations such as
yxyyxxyyyx 232 and 56 32332
are said to be in implicit form （隐式表达式） .
73
Example 26
Find Find dy/dx dy/dx if if 322 xyyx
Solution:Solution:
We are going to differentiate both sides of the given equation with We are going to differentiate both sides of the given equation with respect to respect to xx. Firstly, we temporarily replace . Firstly, we temporarily replace yy by by f(x)f(x) and rewrite and rewrite
the equation as the equation as . . Secondly, we differentiate both Secondly, we differentiate both
sides of this equation term by term with respect to sides of this equation term by term with respect to xx: :
322 ))(()( xxfxfx
)(
2
]))([()]([
22
322
3
22
3)(2)()(
][]))(()([
xdx
d
xfdx
dxfx
dx
d
xdx
dfxfx
dx
dxf
dx
dfx
xdx
dxfxfx
dx
d
To be continued
74
Thus, we have Thus, we have
dx
df
xfx
xxfx
dx
df
xxfxdx
dfxfx
xxfxdx
dfxf
dx
dfx
dx
dfx
dx
dfxfxxf
dx
dfx
for solve )(2
)(23
termscombine )(23)](2[
equation theof side oneon )(23)(2
terms allgather 3)(2)2)((
2
2
22
22
22
Finally, replace f(x) by y to get
yx
xyx
dx
dy
2
232
2
75
Implicit Implicit DifferentiationDifferentiation: Suppose an equation defines : Suppose an equation defines yy implicitly as a differentiable function of implicitly as a differentiable function of xx. To find . To find df/dxdf/dx
1.1. Differentiate both sides of equation with respect to Differentiate both sides of equation with respect to xx. . remember that remember that yy is really a function of is really a function of xx and use the and use the
chain rule when differentiating terms containing chain rule when differentiating terms containing yy..2. Solve the differentiated equation algebraically for 2. Solve the differentiated equation algebraically for dy/dx.dy/dx. Exercise
Find Find dy/dx dy/dx by implicit differentiation where by implicit differentiation where xyyx 33
76
Thus, the slope at Thus, the slope at (3,4)(3,4) is is
The slope at The slope at (3,-4)(3,-4) is is
Differentiating both sides of the equation with respect to Differentiating both sides of the equation with respect to xx, we have , we have
Find the slope of the tangent line to the circle at Find the slope of the tangent line to the circle at
the point the point (3,4).(3,4). What is the slope at the point What is the slope at the point (3,-4)?(3,-4)?
Computing the slope of a tangent line by implicit differentiation
2522 yx
y
x
dx
dy
dx
dyyx 022
4
3
)4,3(
dx
dy
Example 27
Solution:Solution:
4
3
4
3
)4,3(
dx
dy
77
Application to Economics
Example 28
Suppose the output at a certain factory isSuppose the output at a certain factory is units, where units, where xx is the number of hours of skilled labor is the number of hours of skilled labor used and y is the number of hours of unskilled labor. The used and y is the number of hours of unskilled labor. The current labor force consists of current labor force consists of 3030 hours of skill labor and hours of skill labor and 2020 hours of unskilled labor. hours of unskilled labor.
Question: Use calculus to estimate the change in unskilled Question: Use calculus to estimate the change in unskilled labor labor yy that should be made to offset a that should be made to offset a 11-hour increase in -hour increase in skilled labor skilled labor xx so that output will be maintained at its so that output will be maintained at its current level.current level.
3232 yyxxQ
78
Solution:Solution:If output is to be maintained at the current level, which is the value If output is to be maintained at the current level, which is the value of of QQ when when x=30x=30 and and y=20y=20, the relationship between skilled labor , the relationship between skilled labor xx and unskilled labor and unskilled labor y y is given by the equation is given by the equation
3232)20,30(000,80 yyxxQ The goal is to estimate the change in The goal is to estimate the change in yy that corresponds to a that corresponds to a 11-unit -unit increase in increase in xx when when xx and and yy are related by above equation. As we are related by above equation. As we know, the change in know, the change in yy caused by a caused by a 11-unit increase in -unit increase in xx can be can be approximated by the derivative approximated by the derivative dy/dx. dy/dx. Using implicit Using implicit differentiation, we have differentiation, we have
22
2
222
222
3
26
26)3(
3260
yx
xyx
dx
dy
xyxdx
dyyx
dx
dyyxy
dx
dyxx
Now evaluate this derivative when x=30 and y=20 to conclude that Now evaluate this derivative when x=30 and y=20 to conclude that hours 14.3
)20(3)30(
)20)(30(2)30(6in Change
22
2
20
30
y
xdx
dyy
79
In certain practical problemsIn certain practical problems, x and y are related by an equation and can be regarded as a function of a third variable t, which often represents time. Then implicit differentiation can be used to relate dx/dt to dy/dt. This kind of problem is said to involve related rates.
A procedure for solving related rates problemsA procedure for solving related rates problems
1.1. Find a formula relating the variables.Find a formula relating the variables.
2.2. Use implicit differentiation to find how the rates are related.Use implicit differentiation to find how the rates are related.
3.3. Substitute any given numerical information into the equation in Substitute any given numerical information into the equation in step 2 to find the desired rate of change. step 2 to find the desired rate of change.
80
The manager of a company determines that when The manager of a company determines that when q q hundred units of a particular commodity are hundred units of a particular commodity are produced, the cost of production is produced, the cost of production is CC thousand thousand dollars, where . When dollars, where . When 1500 1500 units units are being produced, the level of production is are being produced, the level of production is increasing at the rate of increasing at the rate of 2020 units per week. units per week.
What is the total cost at this time and at what rate What is the total cost at this time and at what rate is it changing?is it changing?
Example 29
42753 32 qC
81
We want to find We want to find dC/dtdC/dt when when q=15q=15 and and dq/dt=0.2dq/dt=0.2. Differentiating . Differentiating the equation implicitly with respect to time, the equation implicitly with respect to time, we get we get
Solution:Solution:
42753 32 qC
0332 2
dt
dqq
dt
dCC
so that so that
dt
dq
C
q
dt
dC
2
9 2
When When q=15q=15, the cost , the cost CC satisfies satisfies
12014400)15(342754275)15(3 3232 CCCand by substituting and by substituting q=15q=15, , C=120C=120 and and dq/dt=0.2dq/dt=0.2 into the formula into the formula for for dC/dtdC/dt, we obtain , we obtain
per week. dollars nd thousa6875.1)2.0()120(2
)15(9 2
dt
dC
82
A storm at sea has damaged an oil rig. Oil spills A storm at sea has damaged an oil rig. Oil spills from the rupture at the constant rate of from the rupture at the constant rate of 6060 , forming a slick that is roughly , forming a slick that is roughly circular in shape and circular in shape and 3 inches3 inches thick. thick.
a.a. How fast is the radius of the slick increasing How fast is the radius of the slick increasing when the radius is when the radius is 70 feet70 feet??
b.b. Suppose the rupture is repaired in such a way that Suppose the rupture is repaired in such a way that flow is shut off instantaneously. If the radius of flow is shut off instantaneously. If the radius of the slick is increasing at the rate of the slick is increasing at the rate of 0.2 ft/min 0.2 ft/min when the flow stops. What is the total volume of when the flow stops. What is the total volume of oil that spilled onto the sea? oil that spilled onto the sea?
Example 30
min/3ft
83
Solution:Solution:
We can think of the slick as a cylinder of oil of radius We can think of the slick as a cylinder of oil of radius rr feet and feet and thickness thickness h=3/12=0.25 feeth=3/12=0.25 feet. Such a cylinder will have volume . Such a cylinder will have volume
322 ft 25.0 rhrV Differentiating implicitly in this equation with respect to time Differentiating implicitly in this equation with respect to time t, t, and since and since dV/dt=60dV/dt=60 at all times, we get at all times, we get
dt
drr
dt
drr
dt
drr
dt
dV 5.0605.0225.0
a. We want to find a. We want to find dr/dtdr/dt when when r=70, r=70, so that so that
ft/min 55.0)70()5.0(
60
dt
dr
b. Since b. Since dr/dt=0.2dr/dt=0.2 at that instant, we have at that instant, we have feet 191)2.0(5.0
60
r
Therefore, the total amount of oil spilled is Therefore, the total amount of oil spilled is 32 ft 28652)191(25.0 V
84
Exercise
A lake is polluted by waste from a plant located on A lake is polluted by waste from a plant located on its shore. Ecologists determine that when the level its shore. Ecologists determine that when the level of pollute is of pollute is xx parts per million (ppm), there will be parts per million (ppm), there will be FF fish of a certain species in the lake, where fish of a certain species in the lake, where
xF
3
32000
When there are When there are 40004000 fish left in the lake, the fish left in the lake, the pollution is increasing at the rate of pollution is increasing at the rate of 1.4 ppm/year1.4 ppm/year. . At what rate is the fish population changing at this At what rate is the fish population changing at this time? time?
85
Summary Definition of the DerivativeDefinition of the Derivative
h
xfhxfxf
h
)()(lim)(
0
Interpretation of the DerivativeInterpretation of the Derivative
Slope as a DerivativeSlope as a Derivative: The slope of the tangent line to : The slope of the tangent line to the curve the curve y=f(x)y=f(x) at the point at the point (c,f(c))(c,f(c)) is is
Instantaneous Rate of Change as a DerivativeInstantaneous Rate of Change as a Derivative: The : The rate of change of rate of change of f(x) f(x) with respect to with respect to x x when when x=cx=c is is given by given by f’(c)f’(c)
)(tan cfm
86
Sign of The DerivativeSign of The Derivative
If the function If the function ff is differentiable at x=c, then is differentiable at x=c, then
ff is increasing at x=c if >0 is increasing at x=c if >0)(cf
ff is decreasing at x=c if <0 is decreasing at x=c if <0)(cf
Techniques of DifferentiationTechniques of Differentiation
0cdx
d 1][ nn nxxdx
d )()( xfdx
dcxcf
dx
d
)]([)]([)]()([ xgdx
dxf
dx
dxgxf
dx
d
)]([)()]([)()()( xfdx
dxgxg
dx
dxfxgxf
dx
d The Product RuleThe Product Rule
0)( if )(
)]([)()]([)(]
)(
)([
2
xg
xg
xgdxd
xfxfdxd
xg
xg
xf
dx
dThe Quotient Rule
87
The Chain RuleThe Chain Rule
dx
du
du
dy
dx
dy )())(( xgxgf
dx
dy
)]([)]([)]([ 1 xhdx
dxhnxh
dx
d nn The General Power Rule
The Higher -order DerivativeThe Higher -order Derivative
Application of DerivativeApplication of Derivative
Tangent line, Rectilinear Motion, Projectile MotionTangent line, Rectilinear Motion, Projectile Motion
)(or 2
2
xfdx
yd
)(or )( xfdx
yd nn
n
The Second Derivative
The nth Derivative
88
The marginal cost is , it approximates ,The marginal cost is , it approximates ,the additional cost generated by producing one more unit.the additional cost generated by producing one more unit.
)( 0xC )()1( 00 xCxC
Marginal Analysis and Approximation by incrementsMarginal Analysis and Approximation by increments
h
xChxCxCxCxC
xCxCh
)()(lim)()()1(
1
)()1( 00
0000
00
xxfxfxxf )()()( 000 Approximation by Increment
Marginal Revenue Marginal Profit
Implicit Differentiation and Related RateImplicit Differentiation and Related Rate
