Lesson 11: Implicit Differentiation

. . . . . .

Section2.6ImplicitDifferentiation

V63.0121.027, CalculusI

October8, 2009

Announcements

I MidtermnextThursday, covering§§1.1–2.4.

..Imagecredit: TelstarLogistics

http://www.flickr.com/photos/telstar/

. . . . . .

Outline

Thebigidea, byexample

ExamplesVerticalandHorizontalTangentsOrthogonalTrajectoriesChemistry

Thepowerruleforrationalpowers

. . . . . .

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.

. . . . . .

MotivatingExample


x2 + y2 = 1


. .x

.y

.Solution(Explicit)



= − −2x2√1− x2

=x√

1− x2

I Evaluate:dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

=34.

. . . . . .

MotivatingExample


x2 + y2 = 1


. .x

.y

.

Solution(Explicit)



= − −2x2√1− x2

=x√

1− x2

I Evaluate:dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

=34.

. . . . . .

MotivatingExample


x2 + y2 = 1


. .x

.y

.Solution(Explicit)



= − −2x2√1− x2

=x√

1− x2

I Evaluate:dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

=34.

. . . . . .

MotivatingExample


x2 + y2 = 1


. .x

.y

.Solution(Explicit)



= − −2x2√1− x2

=x√

1− x2

I Evaluate:dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

=34.

. . . . . .

MotivatingExample


x2 + y2 = 1


. .x

.y

.Solution(Explicit)



= − −2x2√1− x2

=x√

1− x2

I Evaluate:dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

=34.

. . . . . .

MotivatingExample


x2 + y2 = 1


. .x

.y

.Solution(Explicit)



= − −2x2√1− x2

=x√

1− x2

I Evaluate:dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

=34.

. . . . . .

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)

. . . . . .




x2 + (f(x))2 = 1


2x + 2f(x) · f′(x) = 0


f′(x) = − xf(x)

. . . . . .




x2 + (f(x))2 = 1


2x + 2f(x) · f′(x) = 0


f′(x) = − xf(x)

. . . . . .

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

.

. . . . . .

Yes, wecan!





. .x

.y

.

. . . . . .

Yes, wecan!





. .x

.y

.

.lookslikeafunction

. . . . . .

Yes, wecan!





. .x

.y

.

.

. . . . . .

Yes, wecan!





. .x

.y

.

.

. . . . . .

Yes, wecan!





. .x

.y

.

.

.lookslikeafunction

. . . . . .

Yes, wecan!





. .x

.y

.

.

. . . . . .

Yes, wecan!





. .x

.y

.

.

. . . . . .

Yes, wecan!





. .x

.y

.

.

.does not look like afunction, but that’sOK—there are onlytwo points like this

.

. . . . . .

Yes, wecan!





. .x

.y

.

.lookslikeafunction

. . . . . .

Yes, wecan!





. .x

.y

.

.lookslikeafunction

. . . . . .

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.

. . . . . .



Solution


= 0


I Isolate:dydx

= − xy.

I Evaluate:dydx

∣∣∣∣( 35 ,− 4

5)=

3/54/5

=34.

. . . . . .



Solution


= 0


I Isolate:dydx

= − xy.

I Evaluate:dydx

∣∣∣∣( 35 ,− 4

5)=

3/54/5

=34.

. . . . . .



Solution


= 0


I Isolate:dydx

= − xy.

I Evaluate:dydx

∣∣∣∣( 35 ,− 4

5)=

3/54/5

=34.

. . . . . .



Solution


= 0


I Isolate:dydx

= − xy.

I Evaluate:dydx

∣∣∣∣( 35 ,− 4

5)=

3/54/5

=34.

. . . . . .

Summary

Ifarelationisgivenbetween x and y whichisn’tafunction:

I “Mostofthetime”, i.e., “atmostplaces” y canbeassumedtobeafunctionofx

I wemaydifferentiatetherelationasis

I Solvingfordydx

doesgivethe

slopeofthetangentlinetothecurveatapointonthecurve.

. .x

.y

.

. . . . . .

Mnemonic

Explicit Implicit

y = f(x) F(x, y) = k

. . . . . .

Outline




. . . . . .

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).

. . . . . .


y2 = x2(x + 1) = x3 + x2


.

.SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives

2ydydx

= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

dydx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

4.


(x− 3).

. . . . . .


y2 = x2(x + 1) = x3 + x2


.

.SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives

2ydydx

= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

dydx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

4.


(x− 3).

. . . . . .

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.

. . . . . .

ExampleFindthehorizontaltangentlinestothesamecurve: y2 = x3 + x2

SolutionWehavetosolvethesetwoequations:

.

.y2 = x3 + x2

[(x, y) is onthe curve]

.1.3x2 + 2x

2y= 0

[tangent lineis horizontal]

.2

. . . . . .

ExampleFindthehorizontaltangentlinestothesamecurve: y2 = x3 + x2

SolutionWehavetosolvethesetwoequations:

.

.y2 = x3 + x2


.1.3x2 + 2x

2y= 0

[tangent lineis horizontal]

.2

. . . . . .

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.

. . . . . .


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



y2 = (−23)3 + (−2

3)2 =⇒ y = ± 2

3√3

,


. . . . . .


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



y2 = (−23)3 + (−2

3)2 =⇒ y = ± 2

3√3

,


. . . . . .

HorizontalTangents

..

.(−2

3 ,2

3√3

).

.(−2

3 ,−2

3√3

)

.

.node

..(−1, 0)

. . . . . .

HorizontalTangents

..

.(−2

3 ,2

3√3

).

.(−2

3 ,−2

3√3

) .

.node

..(−1, 0)

. . . . . .

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


.1.

2y3x2 + 2x

= 0

[tangent lineis vertical]

.2

. . . . . .


Solution


= 0.


2y = 3x2dxdy

+ 2xdxdy

, sodxdy

=2y



.

.y2 = x3 + x2


.1.

2y3x2 + 2x

= 0


.2

. . . . . .


Solution


= 0.


2y = 3x2dxdy

+ 2xdxdy

, sodxdy

=2y


reciprocalof dy/dx).

I Wemustsolve

.

.y2 = x3 + x2


.1.

2y3x2 + 2x

= 0


.2

. . . . . .


Solution


= 0.


2y = 3x2dxdy

+ 2xdxdy

, sodxdy

=2y



.

.y2 = x3 + x2


.1.

2y3x2 + 2x

= 0


.2

. . . . . .

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.

. . . . . .

Solution, continued


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,


. . . . . .

Solution, continued


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,


. . . . . .

VerticalTangents

.

..(−2

3 ,2

3√3

).

.(−2

3 ,−2

3√3

)

.

.node

..(−1, 0)

. . . . . .

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

. . . . . .

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

. . . . . .

ExamplesExampleShowthatthefamiliesofcurves

xy = c x2 − y2 = k

areorthogonal, thatis, theyintersectatrightangles.

SolutionInthefirstcurve,

y + xy′ = 0 =⇒ y′ = −yx

Inthesecondcurve,

2x− 2yy′ = 0 = =⇒ y′ =xy

Theproductis −1, sothetangentlinesareperpendicularwherevertheyintersect.

. . . . . .

OrthogonalFamiliesofCurves

. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


. .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

. . . . . .


xy = c x2 − y2 = k



y + xy′ = 0 =⇒ y′ = −yx

Inthesecondcurve,

2x− 2yy′ = 0 = =⇒ y′ =xy


. . . . . .


xy = c x2 − y2 = k



y + xy′ = 0 =⇒ y′ = −yx

Inthesecondcurve,

2x− 2yy′ = 0 = =⇒ y′ =xy


. . . . . .

MusicSelection

“TheCurseofCurves”byCuteisWhatWeAimFor

. . . . . .

Idealgases

The idealgaslaw relatestemperature, pressure, andvolumeofagas:

PV = nRT

(R isaconstant, n istheamountofgasinmoles)

.

.Imagecredit: ScottBeale/LaughingSquid

http://www.laughingsquid.com/

. . . . . .

Compressibility

DefinitionThe isothermiccompressibility ofafluidisdefinedby

β = −dVdP

1V

withtemperatureheldconstant.

Approximatelywehave

∆V∆P

≈ dVdP

= −βV =⇒ ∆VV

≈ −β∆P

Thesmallerthe β, the“harder”thefluid.

. . . . . .

Compressibility

DefinitionThe isothermiccompressibility ofafluidisdefinedby

β = −dVdP

1V

withtemperatureheldconstant.

Approximatelywehave

∆V∆P

≈ dVdP

= −βV =⇒ ∆VV

≈ −β∆P

Thesmallerthe β, the“harder”thefluid.

. . . . . .

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.

. . . . . .

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.

. . . . . .

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

http://commons.wikimedia.org/wiki/Image:Vdwaals.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

http://commons.wikimedia.org/wiki/Image:Vdwaals.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?


da?

. . . . . .


P +an2

V2

)dVdP

+ (V− bn)

(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)


I Whatif a = b = 0?


db?


da?

. . . . . .


P +an2

V2

)dVdP

+ (V− bn)

(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)


I Whatif a = b = 0?


db?


da?

. . . . . .


P +an2

V2

)dVdP

+ (V− bn)

(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)


I Whatif a = b = 0?


db?


da?

. . . . . .


P +an2

V2

)dVdP

+ (V− bn)

(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)


I Whatif a = b = 0?


db?


da?

. . . . . .

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).

. . . . . .

Outline




. . . . . .

Usingimplicitdifferentiationtofindderivatives

Example

Finddydx

if y =√x.

SolutionIf y =

√x, then

y2 = x,

so

2ydydx

= 1 =⇒ dydx

=12y

=1

2√x.

. . . . . .

Usingimplicitdifferentiationtofindderivatives

Example

Finddydx

if y =√x.

SolutionIf y =

√x, then

y2 = x,

so

2ydydx

= 1 =⇒ dydx

=12y

=1

2√x.

. . . . . .


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

. . . . . .



pqxp/q−1.

Proof.Wehave


= pxp−1 =⇒ dydx

=pq· x

p−1

yq−1


xp−1

yq−1 = xp−1−(p−p/q) = xp/q−1

. . . . . .



pqxp/q−1.

Proof.Wehave


= pxp−1 =⇒ dydx

=pq· x

p−1

yq−1


xp−1

yq−1 = xp−1−(p−p/q) = xp/q−1

. . . . . .

Summary

Ifarelationisgivenbetween x and y whichisn’tafunction:

I “Mostofthetime”, i.e., “atmostplaces” y canbeassumedtobeafunctionofx

I wemaydifferentiatetherelationasis

I Solvingfordydx

doesgivethe

slopeofthetangentlinetothecurveatapointonthecurve.

. .x

.y

.

Education

Lesson 11: Implicit Differentiation