Derivative 1 of 30 Average Rate of Change (Secant Slope) Definition: The average change (secant slope) of a function over a particular x interval [a, b] or [a, x]. x x f x x f m Δ − Δ + = ) ( ) ( or a x a f x f m − − = ) ( ) ( Example: What is the average rate of change of the function 1 4 ) ( + = x x f over the interval [0, 2]? 1 1 ) 0 ( 4 ) 0 ( = + = f 3 1 ) 2 ( 4 ) 2 ( = + = f 1 0 2 1 3 0 2 ) 0 ( ) 2 ( = − − = − − = f f m The average rate (secant slope) over the interval [0, 2] is 1.
Transcript
of
Derivative 1 of 30
Average Rate of Change (Secant Slope)
Definition: The average change (secant slope) of a function over a particular x interval [a, b] or [a, x].
xxfxxfm
Δ
−Δ+=
)()( or axafxfm
−
−=
)()(
Example: What is the average rate of change of the function
14)( += xxf over the interval [0, 2]?
11)0(4)0( =+=f
31)2(4)2( =+=f
10213
02)0()2(
=−
−=
−
−=
ffm
The average rate (secant slope) over the interval [0, 2] is 1.
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Derivative 2 of 30
Instantaneous Rate of Change (Tangent Slope)
Definition: The change (tangent slope) at a particular point. Here, the x interval (change in x or xΔ ) approaches 0.
hxfhxfm
h
)()(lim0
−+=
→
Example:
What is the instantaneous rate of change of the function 3)( 2 −= xxf at 4=x ?
3)( 2 −= xxf 32)( 22 −++=+ hxhxhxf
8)4(22
2lim2lim
)3()32(lim
)()(lim
0
2
0
222
0
0
===
+=+
=
−−−++=
−+=
→→
→
→
x
hxhhxh
hxhxhx
hxfhxfm
hh
h
h
The instantaneous rate (tangent slope) at 4=x is 8.
slopetangent
slopesecant
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Derivative 3 of 30
Definition of the Derivative
The derivative of a function is the instantaneous rate of change at the x value. Graphically, it is tangent slope of the function.
Definition
hxfhxfxf
h
)()(lim)(0
−+=ʹ′
→
Alternative Definition
axafxfxf
xa −
−=ʹ′
→
)()(lim)(
yʹ′ “y prime” f ʹ′ or )(xf ʹ′ “f prime” y ʹ′ʹ′ “y double prime” y ʹ′ʹ′ʹ′ “y triple prime”
)4(y “4th derivative of y”
dxdy “dy dx” or “the derivative of y with respect to x”
)(xfdxd “d dx of f at x” or the derivative of f at x”
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Derivative 4 of 30
Differentiability
For a function to be differentiable: it must be continuous and its derivative must be continuous. Check the domain of the original function and its derivative.
Occurrences where the Derivative Fails to Exist Discontinuity
Corner: derivative from each side differ
Vertical Tangent: derivative from each side both approaches ∞ or – ∞
Cusps: derivative from one side
approaches ∞ and the other side – ∞
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Derivative 5 of 30
Power Rule
1)(
)(
−ʹ′=ʹ′
=
n
n
nuuxf
uxf
Chain Rule
))(()()(
))(()(
xghxgxf
xghxf
ʹ′ʹ′=ʹ′
=
Take the derivative from Inside Out.Examples
31
2563660
31
2563660)(
31
521230
31
521230)(
61610
6610)(
5 84
58
4
5 332
53
32
252
23
25 2
23
+−+=
+−+=ʹ′ʹ′
++−=
++−=ʹ′
+++=
+++=
−−
−−
−
xxx
xxxxg
xxx
x
xxxxxg
xxxx
xxx
xxg
)35(cos)35sin(40)35(cos4)35sin(10)(
)35(cos)(
)43)(46(4)43(4)46()(
)43()(
232
232
24
32
32
42
−−−=−−−=ʹ′
−=
++=++=ʹ′
+=
xxxxxxxh
xxh
xxxxxxxf
xxxf
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Derivative 6 of 30
Product Rule
uvvuxf
uvxf
ʹ′+ʹ′=ʹ′
=
)(
)(
Quotient Rule
2)(
)(
ggffgxf
gfxf
ʹ′−ʹ′=ʹ′
=
Examples
xxxxxxxxxf
xxxf
5tan65sec15)6(5tan)5sec5(3)(
5tan3)(
22
22
2
+=+=ʹ′
=
xxxxxxx
xxxxxxh
xxxh
5sec5tan5sec255tan5sec155sec3
)5(sec)5tan5sec5)(53()3)(5(sec)(
5sec53)(
2
2
+−=
−−=ʹ′
−=
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Derivative 7 of 30
Implicit Differentiation
Steps 1. Take the derivative (derivative for y is yʹ′ or dy and derivative for x is 1 or dx). 2. Isolate all terms with yʹ′ or dy to one side and everything else to the other.
3. Factor yʹ′ or dy and solve for yʹ′ or dxdy .
Example
yʹ′ method dxdy method
522525)52(255255225522
−+
−−=ʹ′
−−=−+ʹ′−−=ʹ′−ʹ′+ʹ′
+ʹ′=ʹ′++ʹ′++=++
yxyxy
yxyxyyxyyyyx
yyyyyxxxyyxyx
5225
)25()52(255255225522
−+
−−=
−−=−+−−=−+
+=++++=++
yxyx
dxdy
yxdxyxdyydxxdxdxdyydyxdy
dxdyydyydxxdyxdxxyyxyx
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Derivative 8 of 30
Square Root
uuxf
uxf
2)(
)(
ʹ′=ʹ′
=
xxx
xxxxf
xxxf
4323
43246)(
43)(
22
2
+
+=
+
+=ʹ′
+=
Inverse Function
If )(xf has an inverse and baf =)( , then abf =− )(1 .
If )(xf is differentiable at ax = and caf =ʹ′ )( , then c
bf 1)()( 1 =ʹ′− .
Example Find the slope of the inverse function of xxxf += 3)( at )2()( 1 ʹ′−f .
41)1(3)1('13)(211)1(
2
2
3
=+=+=ʹ′=+=
fxxf
f
41)2()(
1)2(1
1
=ʹ′
=
−
−
f
f
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Derivative 9 of 30
Trigonometric
uuxf
uxf
cos)(
sin)(
ʹ′=ʹ′
=
uuxf
uxf
sin)(
cos)(
ʹ′−=ʹ′
=
uuxf
uxf
2sec)(
tan)(
ʹ′=ʹ′
=
uuxf
uxf
2csc)(
cot)(
ʹ′−=ʹ′
=
uuuxf
uxf
tansec)(
sec)(
ʹ′=ʹ′
=
uuuxf
uxf
cotcsc)(
csc)(
ʹ′−=ʹ′
=
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Derivative 10 of 30
Inverse Trigonometric
2
1
1)(
sin)(
u
uxf
uxf
−
ʹ′=ʹ′
= −
2
1
1)(
cos)(
u
uxf
uxf
−
ʹ′−=ʹ′
= −
2
1
1)(
tan)(
uuxf
uxf
+
ʹ′=ʹ′
= −
2
1
1)(
cot)(
uuxf
uxf
+
ʹ′−=ʹ′
= −
1)(
sec)(
2
1
−
ʹ′=ʹ′
= −
uu
uxf
uxf
1)(
csc)(
2
1
−
ʹ′−=ʹ′
= −
uu
uxf
uxf
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Derivative 11 of 30
Logarithmic & Exponential
u
u
euxf
exf
ʹ′=ʹ′
=
)(
)(
uuxf
uxf
ʹ′=ʹ′
=
)(
ln)(
aauxf
axf
u
u
ln)(
)(
ʹ′=ʹ′
=
auuxf
uxf a
ln)(
log)(
ʹ′=ʹ′
=
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Derivative 12 of 30
Piecewise function Take the derivative of each piece separately. Note: The less than or greater than symbols should not have an equals sign after taking the derivative.
Examples
⎩⎨⎧
<−
>+=ʹ′
⎩⎨⎧
≤+−
>=
4,2104,ln33
)(
4,4254,ln3
)( 2
xxxx
xf
xxxxxx
xf
⎪⎩
⎪⎨
⎧
>−
<<
<
=ʹ′
⎪⎩
⎪⎨
⎧
>
≤<
≤+
=
5,csc55.2,05.2,8
)(
5,cot55.2,55.2,54
)(
2
2
xxx
xxxg
xxx
xxxg
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Derivative 13 of 30
Absolute Value Rewrite as piecewise function. Solve for x inside the absolute value symbol. The inside of the absolute value should be positive with >x and negative with <x . If quadratic, the inside of the absolute value should be negative with the numbers in between. Then, follow the steps for a piecewise function to take the derivative.
Examples
⎩⎨⎧
−<−
−>=ʹ′
⎩⎨⎧
−<−−
−≥+=
⎩⎨⎧
−<−+−
−≥−+=
−+=
3,23,2
)(
3,1023,22
3,4)62(3,462
462)(
xx
xf
xxxx
xxxx
xxf
⎩⎨⎧
<<−+−
−<>−=ʹ′
⎩⎨⎧
<<−++−
−≤≥−−=
⎩⎨⎧
<<−+−−−
−≤≥+−−=
+−−=
25.1,145.1or 2,14
)(
25.1,1125.1or 2,12
25.1,5)62(5.1or 2,562
562)(
2
2
2
2
2
xxxxx
xf
xxxxxxx
xxxxxxx
xxxf
3062−=
=+x
x2 ,5.1
0)32)(2(062
−==+−=−−
xxxxx
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Derivative 14 of 30
Extremas – Local (Relative) or Absolute Maximum and Minimums
Critical Point for Extremas Critical Points for Points of Inflection
End Points 0)( =ʹ′ʹ′ xf
0)( =ʹ′ xf DNExf =ʹ′ʹ′ )(
DNExf =ʹ′ )(
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Derivative 15 of 30
8− 8
1st Derivative Line Test Use the first derivative line test to determine extremas (maximums and minimums) and intervals of increasing and decreasing values (slope).
Steps 1. Take 1st derivative, find critical points, and plot them on a number line. 2. Use x values between to find intervals the function is increasing or decreasing. 3. Sketch the graph, determine extremes and find the y values.
Example:
2
2
2
8
82)(
8)(
xxxf
xxxf
−
+−=ʹ′
−=
End Points: 88 08 2 ≤≤−⇒≥− xx
DNExf =ʹ′ )( : 88 08 2 <<−⇒>− xx
0)( =ʹ′ xf : 2 ,2 082 2 −=⇒=+− xx Abs. Max is 4 at 2=x Abs. Min is – 4 at 2−=x Local Max is 0 at 8−=x Local Min is 0 at 8=x Increasing on the interval )2 ,2(−
Decreasing on the intervals )8 ,2()2 ,8( ∪−−
0)8(4)2(4)2(0)8(
=
=−=−
=−
ffff
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Derivative 16 of 30
2nd Derivative Line Test
Use the second derivative line test to determine intervals of concavity (concave up or down) and point of inflections (change in concavity).
Steps 1. Take 2nd derivative, find critical point, and plot them on a number line. 2. Use x values between to find intervals the function is concave up or down. 3. Determine points of inflection and find the y values.
Example:
xx
xx
x
exexfexexf
xexf
2)()()(
+=ʹ′ʹ′+=ʹ′
=
DNExf =ʹ′ʹ′ )( : 2nd derivative is continuous
0)( =ʹ′ʹ′ xf : 2 0)2( −=⇒=+ xxex
Point of Inflection is )27.0 ,2( −− Concave Up on the interval ) ,2( ∞− Concave Down on the interval )2 ,( −−∞
27.0)2( −=−f
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Derivative 17 of 30
2nd Derivative Test for Extremas
If the function is continuous and differentiable, use the second derivative test for extremas to find maximums and minimums.
Cases for Extremas
If 0)( =ʹ′ cf and 0)( <ʹ′ʹ′ cf , then f has a local maximum at cx = .
If 0)( =ʹ′ cf and 0)( >ʹ′ʹ′ cf , then f has a local minimum at cx = .
Example
xxfxxf
xxxf
6)(153)(
512)(2
3
=ʹ′ʹ′−=ʹ′
−−=
0)( =ʹ′ xf : 2 ,2 )4(30 2 −=⇒−= xx
12)2(6)2(12)2(6)2(
==ʹ′ʹ′−=−=−ʹ′ʹ′
ff
Local Max is 11 at 2−=x Local Min is –21 at 2=x
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Derivative 18 of 30
Interpreting the graph of f
)(xf : x intercepts are the Zeros of the Function.
)(xf ʹ′ : Slope increases (or decreases). Hills (or valleys) are Max (or Min).
)(xf ʹ′ʹ′ : Curvature is Concavity. Change in curvature is Point of the Inflection.
Example
)(xf : Zeroes are 2 ,0 ,1−=x )(xf ʹ′
Local Max is 0.6 at 6.0−=x Local Min is –2.1 at 2.1=x Increasing on the interval ) ,2.1()6.0 ,( ∞∪−−∞ Decreasing on the interval )2.1 ,6.0(−
)(xf ʹ′ʹ′ Point of Inflection is –1 at 4.0=x Concave Up on the interval ) ,4.0( ∞ Concave Down on the interval )4.0 ,(−∞
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Derivative 19 of 30
Interpreting the graph of f ʹ′
)(xf ʹ′ : Positive and negative y-values are Slopes. x-intercepts are Max (or Min).
)(xf ʹ′ʹ′ : Slope of graph is Concavity. Hills (or valleys) are Points of the Inflection.
Note: Use the 1st and 2nd Derivative Line Test to understand the original function.
Example
)(xf ʹ′
Local Max at 1=x Local Max at 3=x Increasing on the interval ) ,3()1 ,( ∞∪−∞ Decreasing on the interval )3 ,1(
)(xf ʹ′ʹ′ Point of Inflection at 2 ,5.0 ,2 −−=x Concave Up on the interval ) ,2()5.0 ,2( ∞∪−− Concave Down on the interval )2 ,5.0()2 ,( −∪−−∞
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Derivative 20 of 30
Interpreting a Table
The best way to interpret a table of 1st and 2nd derivatives is to perform the 1st and 2nd Derivative Line Tests. This will help to make conclusion about the original function.
Example f is a continuous function with the following properties. Determine the behavior.
x 2<x 2 42 << x 4 64 << x 6 6>x f 1 4 7 f ʹ′ negative 0 positive positive 0 negative f ʹ′ʹ′ positive positive 0 negative negative
Local Max is 7 at 6=x Local Min is 1 at 2=x Increasing on the interval )6 ,2( Decreasing on the interval ) ,6()2 ,( ∞∪−∞ Point of inflection is )4 ,4( Concave up on the interval )4 ,(−∞ Concave Down on the interval ) ,4( ∞
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Derivative 21 of 30
Odd Functions
Properties )()( xfxf −=−
Symmetry about the origin 1st derivative will be same 2nd derivative will be opposite
)0 ,0( is a point on the graph
Even Functions
Properties )()( xfxf =−
Symmetry about the y-axis 1st derivative will be opposite 2nd derivative will be same
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Derivative 22 of 30
Sketching the Original Function from 1st & 2nd Derivative Test
Steps 1. Plot any points given. 2. Sketch with a dashed line using the 1st derivative line test. 3. Following the dashed line, use a solid line showing the curvature of the graph
based on the 2nd derivative line test.
Example
4)1(2)2(
==−
ff 1)4(
1)3(=−=
ff
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Derivative 23 of 30
Mean Value Theorem
If a function is continuous at every point of the close interval ] ,[ ba and differential at every point of its interior, the open interval ) ,( ba , then there is at least one point c in ) ,( ba whose tangent slope is equal to the average slope from ] ,[ ba .
abafbfcf
−
−=ʹ′
=)()(
)(
slopesecant slopetangent
Example Find the mean value on the interval
]4 ,0[ of 12)( 2 −+= xxxf .
23)4(1)0(22)(22)(
=−=+=ʹ′+=ʹ′
ff
ccfxxf
2622
04)1(23
22
)()()(
==+
−
−−=+
−
−=ʹ′
cc
c
abafbfcf
7)2( =f
The mean value on the interval ]4 ,0[ is 7 at 2=x .
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Derivative 24 of 30
Linearization If f is differentiable at ax = , then the equation of the tangent line,
11 )()( yxxmxL +−= , where )(afm ʹ′= and )(1 afy =
defines the linearization of f at a. The approximation )()( xLxf ≈ is the standard linear approximation of f at a. The point ax = is the center of the approximation.
Example Find the approximate value of cos (1.75)
without a calculator using 2π
=x as the
center of the approximation.
12
sin)(
02
cos)(
−=⎟⎠
⎞⎜⎝
⎛ʹ′
−=ʹ′
=⎟⎠
⎞⎜⎝
⎛=
π
π
f
xxf
f
xxf
18.02
75.1)75.1(
2)(
02
1)(
)()( 11
−≈+−=
+−=
+⎟⎠
⎞⎜⎝
⎛−−=
+−=
π
π
π
L
xxL
xxL
yxxmxL
The value of cos (1.75) is about – 0.18.
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Derivative 25 of 30
Differentials
Differentials are used to estimate approximate change of a function calculated based on slope where dx is the change in the x variable. Thus, the differential dy or df is:
dxxfdfdxxfdy)()(
ʹ′=ʹ′=
Example
Inflating a spherical balloon changes its radius from 6 inches to 6.1 inches. Use differentials to estimate the change in the volume. Compare it with the true change.
ππππ
π
π
64.1428864.30264.302)1.6(
288)6(
34)( 3
=−=Δ==
=
VVV
rrV
The true change in volume is 3cm 64.14 π .
πππ
4.14)1.0()6(4r 4
2
2
===
dVdrdV
The estimated change in volume is 3cm 4.14 π . The difference (or error)
Example A particle moves along the x-axis. Its position is given by 2cossin)( ++= ttttx . For what values of t, 50 << t , is the particle moving right? Find the total distance traveled by the particle over the interval )5 ,0( .
ttttatt
tttttvttttx
cossin)(cos
sinsincos)(2cossin)(
+−==
−+=++=
71.4 ,57.1 ,0cos0
==t
tt
A particle moves to the right when it’s derivative is positive, which is from )5 ,71.4()1.57 ,0( ∪ . The total distance the particle traveled is 7.05.
05.720.028.657.0
51.2)5(71.2)71.4(57.3)57.1(3)0(
−=−=
==
xxxx
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Derivative 27 of 30
Modeling and Optimization
Steps 1. Sketch a diagram. 2. Write down all clues, questions, formulas. 3. Solve by finding minimum or maximum. 4. Check your answer. Does it make sense?
Example You are designing a rectangular poster to contain 2in 50 of printing with a 4-in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper?
10550
W5050A
==
=
==
L
L
WLWL
2
2008
822008
325048503248
8)4)(LW(A
WA
WWA
WW
WWA
LWWLA
−=ʹ′
++=
+++=
+++=++=
5
20080 2
±=
−=
WW
in 18by in 9
18894
=+=+
LW
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Derivative 28 of 30
Cost & Profit
Derivates can be used to find the extra revenue )(xr , cost )(xc , or profit )(xp resulting from selling or producing one more item x . This is called marginal revenue )(xr ʹ′ , marginal cost )(xcʹ′ , or marginal profit )(xpʹ′ .
)(xr = the revenue from selling x items )(xc = the cost of producing x items )(xp = )(xr – )(xc = the profit from selling x items
A company breaks even when the profit )(xp equals zero.
Example Suppose xxr 8)( = represents revenue and 22)( xxc = represents cost, where x measured in thousands of units. Is there a production level that maximizes profit? If so, what is it?
xx
xp
xxxcxrxp
44)(
28)()()(2
−=ʹ′
−=
−=
1
440
=
−=
x
xx 1000 units
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Derivative 29 of 30
Related Rates
Are equations involving two or more variables that are differentiable function of time t. To solve, write an equation to show the relationship between the two variables. Differentiate the equation with respect to time.
Example Water runs into a conical tank at the rate of 9ft3/min. The tank stands point down and has a height of 10ft and a base radius of 5ft. How fast is the water level rising when the water is 6ft deep?
dtdhh
dtdV
h
hh
hrV
2
3
2
2
41
121
2
31
31
π
π
π
π
=
=
⎟⎠
⎞⎜⎝
⎛=
=
The water level rises about 0.32 ft/min as the tank fills at a rate of 9ft3/min.
2
105 hr
hr
=⇒=
32.01
99
(6) 419 2
≈=
=
=
π
π
π
dtdh
dtdh
dtdh
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Derivative 30 of 30
L’Hôpital’s Rule
If by using direct substitution the limit equals an indeterminate form
∞
∞=
→or
00
)()(limxgxf
ax,
then take the limit as ax→ of the derivate of the top and bottom until the limit exists.
