Lesson 11: Implicit Differentiation
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![Page 1: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/1.jpg)
. . . . . .
Section2.6ImplicitDifferentiation
V63.0121.027, CalculusI
October8, 2009
Announcements
I MidtermnextThursday, covering§§1.1–2.4.
..Imagecredit: TelstarLogistics
![Page 2: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/2.jpg)
. . . . . .
Outline
Thebigidea, byexample
ExamplesVerticalandHorizontalTangentsOrthogonalTrajectoriesChemistry
Thepowerruleforrationalpowers
![Page 3: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/3.jpg)
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
![Page 4: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/4.jpg)
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
![Page 5: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/5.jpg)
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.
Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
![Page 6: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/6.jpg)
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
![Page 7: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/7.jpg)
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
![Page 8: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/8.jpg)
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
![Page 9: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/9.jpg)
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
![Page 10: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/10.jpg)
. . . . . .
MotivatingExample, anotherway
Weknowthat x2 + y2 = 1 doesnotdefine y asafunctionof x,butsupposeitdid.
I Supposewehad y = f(x), sothat
x2 + (f(x))2 = 1
I Wecoulddifferentiatethisequationtoget
2x + 2f(x) · f′(x) = 0
I Wecouldthensolvetoget
f′(x) = − xf(x)
![Page 11: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/11.jpg)
. . . . . .
MotivatingExample, anotherway
Weknowthat x2 + y2 = 1 doesnotdefine y asafunctionof x,butsupposeitdid.
I Supposewehad y = f(x), sothat
x2 + (f(x))2 = 1
I Wecoulddifferentiatethisequationtoget
2x + 2f(x) · f′(x) = 0
I Wecouldthensolvetoget
f′(x) = − xf(x)
![Page 12: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/12.jpg)
. . . . . .
MotivatingExample, anotherway
Weknowthat x2 + y2 = 1 doesnotdefine y asafunctionof x,butsupposeitdid.
I Supposewehad y = f(x), sothat
x2 + (f(x))2 = 1
I Wecoulddifferentiatethisequationtoget
2x + 2f(x) · f′(x) = 0
I Wecouldthensolvetoget
f′(x) = − xf(x)
![Page 13: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/13.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
![Page 14: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/14.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
![Page 15: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/15.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.lookslikeafunction
![Page 16: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/16.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
![Page 17: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/17.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
![Page 18: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/18.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
.lookslikeafunction
![Page 19: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/19.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
![Page 20: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/20.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
![Page 21: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/21.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
.does not look like afunction, but that’sOK—there are onlytwo points like this
.
![Page 22: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/22.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.lookslikeafunction
![Page 23: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/23.jpg)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.lookslikeafunction
![Page 24: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/24.jpg)
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x + 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,− 4
5)=
3/54/5
=34.
![Page 25: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/25.jpg)
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x + 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,− 4
5)=
3/54/5
=34.
![Page 26: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/26.jpg)
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x + 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,− 4
5)=
3/54/5
=34.
![Page 27: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/27.jpg)
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x + 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,− 4
5)=
3/54/5
=34.
![Page 28: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/28.jpg)
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x + 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,− 4
5)=
3/54/5
=34.
![Page 29: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/29.jpg)
. . . . . .
Summary
Ifarelationisgivenbetween x and y whichisn’tafunction:
I “Mostofthetime”, i.e., “atmostplaces” y canbeassumedtobeafunctionofx
I wemaydifferentiatetherelationasis
I Solvingfordydx
doesgivethe
slopeofthetangentlinetothecurveatapointonthecurve.
. .x
.y
.
![Page 30: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/30.jpg)
. . . . . .
Mnemonic
Explicit Implicit
y = f(x) F(x, y) = k
![Page 31: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/31.jpg)
. . . . . .
Outline
Thebigidea, byexample
ExamplesVerticalandHorizontalTangentsOrthogonalTrajectoriesChemistry
Thepowerruleforrationalpowers
![Page 32: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/32.jpg)
. . . . . .
ExampleFindtheequationofthelinetangenttothecurve
y2 = x2(x + 1) = x3 + x2
atthepoint (3,−6).
.
.
SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thustheequationofthetangentlineis y + 6 = −114
(x− 3).
![Page 33: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/33.jpg)
. . . . . .
ExampleFindtheequationofthelinetangenttothecurve
y2 = x2(x + 1) = x3 + x2
atthepoint (3,−6).
.
.SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thustheequationofthetangentlineis y + 6 = −114
(x− 3).
![Page 34: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/34.jpg)
. . . . . .
ExampleFindtheequationofthelinetangenttothecurve
y2 = x2(x + 1) = x3 + x2
atthepoint (3,−6).
.
.SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thustheequationofthetangentlineis y + 6 = −114
(x− 3).
![Page 35: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/35.jpg)
. . . . . .
Lineequationforms
I slope-interceptform
y = mx + b
wheretheslopeis m and (0,b) isontheline.I point-slopeform
y− y0 = m(x− x0)
wheretheslopeis m and (x0, y0) isontheline.
![Page 36: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/36.jpg)
. . . . . .
ExampleFindthehorizontaltangentlinestothesamecurve: y2 = x3 + x2
SolutionWehavetosolvethesetwoequations:
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.3x2 + 2x
2y= 0
[tangent lineis horizontal]
.2
![Page 37: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/37.jpg)
. . . . . .
ExampleFindthehorizontaltangentlinestothesamecurve: y2 = x3 + x2
SolutionWehavetosolvethesetwoequations:
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.3x2 + 2x
2y= 0
[tangent lineis horizontal]
.2
![Page 38: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/38.jpg)
. . . . . .
Solution, continuedI Solvingthesecondequationgives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
(aslongas y ̸= 0). So x = 0 or 3x + 2 = 0.
I Substituting x = 0 intothe first equationgives
y2 = 03 + 02 = 0 =⇒ y = 0
whichwe’vedisallowed. Sonohorizontaltangentsdownthatroad.
I Substituting x = −2/3 intothefirstequationgives
y2 = (−23)3 + (−2
3)2 =⇒ y = ± 2
3√3
,
sotherearetwohorizontaltangents.
![Page 39: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/39.jpg)
. . . . . .
Solution, continuedI Solvingthesecondequationgives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
(aslongas y ̸= 0). So x = 0 or 3x + 2 = 0.I Substituting x = 0 intothe first equationgives
y2 = 03 + 02 = 0 =⇒ y = 0
whichwe’vedisallowed. Sonohorizontaltangentsdownthatroad.
I Substituting x = −2/3 intothefirstequationgives
y2 = (−23)3 + (−2
3)2 =⇒ y = ± 2
3√3
,
sotherearetwohorizontaltangents.
![Page 40: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/40.jpg)
. . . . . .
Solution, continuedI Solvingthesecondequationgives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
(aslongas y ̸= 0). So x = 0 or 3x + 2 = 0.I Substituting x = 0 intothe first equationgives
y2 = 03 + 02 = 0 =⇒ y = 0
whichwe’vedisallowed. Sonohorizontaltangentsdownthatroad.
I Substituting x = −2/3 intothefirstequationgives
y2 = (−23)3 + (−2
3)2 =⇒ y = ± 2
3√3
,
sotherearetwohorizontaltangents.
![Page 41: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/41.jpg)
. . . . . .
HorizontalTangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
)
.
.node
..(−1, 0)
![Page 42: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/42.jpg)
. . . . . .
HorizontalTangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
) .
.node
..(−1, 0)
![Page 43: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/43.jpg)
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
![Page 44: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/44.jpg)
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
![Page 45: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/45.jpg)
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).
I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
![Page 46: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/46.jpg)
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
![Page 47: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/47.jpg)
. . . . . .
Solution, continued
I Solvingthesecondequationgives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(aslongas 3x2 + 2x ̸= 0).
I Substituting y = 0 intothe first equationgives
0 = x3 + x2 = x2(x + 1)
So x = 0 or x = −1.I x = 0 isnotallowedbythefirstequation, but
dxdy
∣∣∣∣(−1,0)
= 0,
sohereisaverticaltangent.
![Page 48: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/48.jpg)
. . . . . .
Solution, continued
I Solvingthesecondequationgives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(aslongas 3x2 + 2x ̸= 0).I Substituting y = 0 intothe first equationgives
0 = x3 + x2 = x2(x + 1)
So x = 0 or x = −1.
I x = 0 isnotallowedbythefirstequation, but
dxdy
∣∣∣∣(−1,0)
= 0,
sohereisaverticaltangent.
![Page 49: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/49.jpg)
. . . . . .
Solution, continued
I Solvingthesecondequationgives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(aslongas 3x2 + 2x ̸= 0).I Substituting y = 0 intothe first equationgives
0 = x3 + x2 = x2(x + 1)
So x = 0 or x = −1.I x = 0 isnotallowedbythefirstequation, but
dxdy
∣∣∣∣(−1,0)
= 0,
sohereisaverticaltangent.
![Page 50: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/50.jpg)
. . . . . .
VerticalTangents
.
..(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
)
.
.node
..(−1, 0)
![Page 51: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/51.jpg)
. . . . . .
ExampleFind y′ if y5 + x2y3 = 1 + y sin(x2).
SolutionDifferentiatingimplicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Collectalltermswith y′ ononesideandalltermswithout y′ ontheother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Nowfactoranddivide:
y′ =2xy(cos x2 − y2)
5y4 + 3x2y2 − sin x2
![Page 52: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/52.jpg)
. . . . . .
ExampleFind y′ if y5 + x2y3 = 1 + y sin(x2).
SolutionDifferentiatingimplicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Collectalltermswith y′ ononesideandalltermswithout y′ ontheother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Nowfactoranddivide:
y′ =2xy(cos x2 − y2)
5y4 + 3x2y2 − sin x2
![Page 53: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/53.jpg)
. . . . . .
ExamplesExampleShowthatthefamiliesofcurves
xy = c x2 − y2 = k
areorthogonal, thatis, theyintersectatrightangles.
SolutionInthefirstcurve,
y + xy′ = 0 =⇒ y′ = −yx
Inthesecondcurve,
2x− 2yy′ = 0 = =⇒ y′ =xy
Theproductis −1, sothetangentlinesareperpendicularwherevertheyintersect.
![Page 54: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/54.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 55: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/55.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 56: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/56.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 57: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/57.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 58: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/58.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 59: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/59.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 60: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/60.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 61: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/61.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 62: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/62.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 63: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/63.jpg)
. . . . . .
OrthogonalFamiliesofCurves
. .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
![Page 64: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/64.jpg)
. . . . . .
ExamplesExampleShowthatthefamiliesofcurves
xy = c x2 − y2 = k
areorthogonal, thatis, theyintersectatrightangles.
SolutionInthefirstcurve,
y + xy′ = 0 =⇒ y′ = −yx
Inthesecondcurve,
2x− 2yy′ = 0 = =⇒ y′ =xy
Theproductis −1, sothetangentlinesareperpendicularwherevertheyintersect.
![Page 65: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/65.jpg)
. . . . . .
ExamplesExampleShowthatthefamiliesofcurves
xy = c x2 − y2 = k
areorthogonal, thatis, theyintersectatrightangles.
SolutionInthefirstcurve,
y + xy′ = 0 =⇒ y′ = −yx
Inthesecondcurve,
2x− 2yy′ = 0 = =⇒ y′ =xy
Theproductis −1, sothetangentlinesareperpendicularwherevertheyintersect.
![Page 66: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/66.jpg)
. . . . . .
MusicSelection
“TheCurseofCurves”byCuteisWhatWeAimFor
![Page 67: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/67.jpg)
. . . . . .
Idealgases
The idealgaslaw relatestemperature, pressure, andvolumeofagas:
PV = nRT
(R isaconstant, n istheamountofgasinmoles)
.
.Imagecredit: ScottBeale/LaughingSquid
![Page 68: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/68.jpg)
. . . . . .
Compressibility
DefinitionThe isothermiccompressibility ofafluidisdefinedby
β = −dVdP
1V
withtemperatureheldconstant.
Approximatelywehave
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
Thesmallerthe β, the“harder”thefluid.
![Page 69: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/69.jpg)
. . . . . .
Compressibility
DefinitionThe isothermiccompressibility ofafluidisdefinedby
β = −dVdP
1V
withtemperatureheldconstant.
Approximatelywehave
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
Thesmallerthe β, the“harder”thefluid.
![Page 70: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/70.jpg)
. . . . . .
ExampleFindtheisothermiccompressibilityofanidealgas.
SolutionIf PV = k (n isconstantforourpurposes, T isconstantbecauseoftheword isothermic, and R reallyisconstant), then
dPdP
· V + PdVdP
= 0 =⇒ dVdP
= −VP
So
β = −1V· dVdP
=1P
Compressibilityandpressureareinverselyrelated.
![Page 71: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/71.jpg)
. . . . . .
ExampleFindtheisothermiccompressibilityofanidealgas.
SolutionIf PV = k (n isconstantforourpurposes, T isconstantbecauseoftheword isothermic, and R reallyisconstant), then
dPdP
· V + PdVdP
= 0 =⇒ dVdP
= −VP
So
β = −1V· dVdP
=1P
Compressibilityandpressureareinverselyrelated.
![Page 72: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/72.jpg)
. . . . . .
NonidealgassesNotthatthere’sanythingwrongwiththat
ExampleThe vanderWaalsequationmakesfewersimplifications:(P + a
n2
V2
)(V− nb) = nRT,
where P isthepressure, V thevolume, T thetemperature, nthenumberofmolesofthegas, R aconstant, a isameasureofattractionbetweenparticlesofthegas,and b ameasureofparticlesize.
...Oxygen
..H
..H
..Oxygen
..H
..H
..Oxygen ..H
..H
.
.
.Hydrogenbonds
..Imagecredit: WikimediaCommons
![Page 73: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/73.jpg)
. . . . . .
NonidealgassesNotthatthere’sanythingwrongwiththat
ExampleThe vanderWaalsequationmakesfewersimplifications:(P + a
n2
V2
)(V− nb) = nRT,
where P isthepressure, V thevolume, T thetemperature, nthenumberofmolesofthegas, R aconstant, a isameasureofattractionbetweenparticlesofthegas,and b ameasureofparticlesize. .
.Imagecredit: WikimediaCommons
![Page 74: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/74.jpg)
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P +an2
V2
)dVdP
+ (V− bn)
(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V + PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβ
db?
I Withouttakingthederivative, whatisthesignofdβ
da?
![Page 75: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/75.jpg)
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P +an2
V2
)dVdP
+ (V− bn)
(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V + PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβ
db?
I Withouttakingthederivative, whatisthesignofdβ
da?
![Page 76: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/76.jpg)
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P +an2
V2
)dVdP
+ (V− bn)
(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V + PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβ
db?
I Withouttakingthederivative, whatisthesignofdβ
da?
![Page 77: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/77.jpg)
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P +an2
V2
)dVdP
+ (V− bn)
(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V + PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβ
db?
I Withouttakingthederivative, whatisthesignofdβ
da?
![Page 78: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/78.jpg)
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P +an2
V2
)dVdP
+ (V− bn)
(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V + PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβ
db?
I Withouttakingthederivative, whatisthesignofdβ
da?
![Page 79: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/79.jpg)
. . . . . .
Nastyderivatives
I
dβ
db= −(2abn3 − an2V + PV3)(nV2) − (nbV2 − V3)(2an3)
(2abn3 − an2V + PV3)2
= −nV3
(an2 + PV2
)(PV3 + an2(2bn− V)
)2 < 0
Idβ
da=
n2(bn− V)(2bn− V)V2(PV3 + an2(2bn− V)
)2 > 0
(aslongas V > 2nb, andit’sprobablytruethat V ≫ 2nb).
![Page 80: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/80.jpg)
. . . . . .
Outline
Thebigidea, byexample
ExamplesVerticalandHorizontalTangentsOrthogonalTrajectoriesChemistry
Thepowerruleforrationalpowers
![Page 81: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/81.jpg)
. . . . . .
Usingimplicitdifferentiationtofindderivatives
Example
Finddydx
if y =√x.
SolutionIf y =
√x, then
y2 = x,
so
2ydydx
= 1 =⇒ dydx
=12y
=1
2√x.
![Page 82: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/82.jpg)
. . . . . .
Usingimplicitdifferentiationtofindderivatives
Example
Finddydx
if y =√x.
SolutionIf y =
√x, then
y2 = x,
so
2ydydx
= 1 =⇒ dydx
=12y
=1
2√x.
![Page 83: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/83.jpg)
. . . . . .
Thepowerruleforrationalpowers
TheoremIf y = xp/q, where p and q areintegers, then y′ =
pqxp/q−1.
Proof.Wehave
yq = xp =⇒ qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
Now yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 = xp−1−(p−p/q) = xp/q−1
![Page 84: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/84.jpg)
. . . . . .
Thepowerruleforrationalpowers
TheoremIf y = xp/q, where p and q areintegers, then y′ =
pqxp/q−1.
Proof.Wehave
yq = xp =⇒ qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
Now yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 = xp−1−(p−p/q) = xp/q−1
![Page 85: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/85.jpg)
. . . . . .
Thepowerruleforrationalpowers
TheoremIf y = xp/q, where p and q areintegers, then y′ =
pqxp/q−1.
Proof.Wehave
yq = xp =⇒ qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
Now yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 = xp−1−(p−p/q) = xp/q−1
![Page 86: Lesson 11: Implicit Differentiation](https://reader034.fdocuments.in/reader034/viewer/2022042700/554a5705b4c905572f8b4cae/html5/thumbnails/86.jpg)
. . . . . .
Summary
Ifarelationisgivenbetween x and y whichisn’tafunction:
I “Mostofthetime”, i.e., “atmostplaces” y canbeassumedtobeafunctionofx
I wemaydifferentiatetherelationasis
I Solvingfordydx
doesgivethe
slopeofthetangentlinetothecurveatapointonthecurve.
. .x
.y
.