Download pdf - Lesson 11: Implicit Differentiation (Section 41 slides)

Section 2.6Implicit Differentiation

V63.0121.041, Calculus I

New York University

October 13, 2010

Announcements

I Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2I Midterm next week. Covers §§1.1–2.5

. . . . . .

. . . . . .

Announcements

I Quiz 2 in recitation thisweek. Covers §§1.5, 1.6,2.1, 2.2

I Midterm next week.Covers §§1.1–2.5

V63.0121.041, Calculus I (NYU) Section 2.6 Implicit Differentiation October 13, 2010 2 / 34

. . . . . .

Objectives

I Use implicit differentationto find the derivative of afunction defined implicitly.


. . . . . .

Outline

The big idea, by example

ExamplesBasic ExamplesVertical and Horizontal TangentsOrthogonal TrajectoriesChemistry

The power rule for rational powers


. . . . . .

Motivating Example

ProblemFind the slope of the linewhich is tangent to thecurve

x2 + y2 = 1

at the point (3/5,−4/5).

. .x

.y

.

Solution (Explicit)

I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)

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.


. . . . . .

Motivating Example


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.


. . . . . .

Motivating Example


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.


. . . . . .

Motivating Example


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.


. . . . . .

Motivating Example


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.


. . . . . .

Motivating Example


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.


. . . . . .

Motivating Example


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.


. . . . . .

Motivating Example, another way

We know that x2 + y2 = 1 does not define y as a function of x, butsuppose it did.

I Suppose we had y = f(x), so that

x2 + (f(x))2 = 1

I We could differentiate this equation to get

2x+ 2f(x) · f′(x) = 0

I We could then solve to get

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, we can!

The beautiful fact (i.e., deep theorem) is that this works!

I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.

I So f(x) is defined “locally”,almost everywhere and isdifferentiable

I The chain rule then appliesfor this local choice.

. .x

.y

.


. . . . . .

Yes, we can!





. .x

.y

.


. . . . . .

Yes, we can!





. .x

.y

.

.looks like a function


. . . . . .

Yes, we can!





. .x

.y

.

.


. . . . . .

Yes, we can!





. .x

.y

.

.


. . . . . .

Yes, we can!





. .x

.y

.

.



. . . . . .

Yes, we can!





. .x

.y

.

.


. . . . . .

Yes, we can!





. .x

.y

.

.


. . . . . .

Yes, we can!





. .x

.y

.

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

.


. . . . . .

Yes, we can!





. .x

.y

.



. . . . . .

Yes, we can!





. .x

.y

.



. . . . . .

Motivating Example, again, with Leibniz notation

ProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).

Solution

I Differentiate: 2x+ 2ydydx

= 0Remember y is assumed to be a function of x!

I Isolate:dydx

= −xy.

I Evaluate:dydx

∣∣∣∣( 35 ,−

45)

=3/54/5

=34.


. . . . . .



Solution


= 0

Remember y is assumed to be a function of x!

I Isolate:dydx

= −xy.

I Evaluate:dydx

∣∣∣∣( 35 ,−

45)

=3/54/5

=34.


. . . . . .



Solution



I Isolate:dydx

= −xy.

I Evaluate:dydx

∣∣∣∣( 35 ,−

45)

=3/54/5

=34.


. . . . . .



Solution



I Isolate:dydx

= −xy.

I Evaluate:dydx

∣∣∣∣( 35 ,−

45)

=3/54/5

=34.


. . . . . .



Solution



I Isolate:dydx

= −xy.

I Evaluate:dydx

∣∣∣∣( 35 ,−

45)

=3/54/5

=34.


. . . . . .

Summary

If a relation is given between x and y which isn’t a function:

I “Most of the time”, i.e., “atmost places” y can beassumed to be a function of x

I we may differentiate therelation as is

I Solving fordydx

does give theslope of the tangent line to thecurve at a point on the curve.

. .x

.y

.


. . . . . .

Outline





. . . . . .

Another Example

Example

Find y′ along the curve y3 + 4xy = x2 + 3.

SolutionImplicitly differentiating, we have

3y2y′ + 4(1 · y+ x · y′) = 2x

Solving for y′ gives

3y2y′ + 4xy′ = 2x− 4y

(3y2 + 4x)y′ = 2x− 4y

=⇒ y′ =2x− 4y3y2 + 4x


. . . . . .

Another Example

Example



3y2y′ + 4(1 · y+ x · y′) = 2x


3y2y′ + 4xy′ = 2x− 4y

(3y2 + 4x)y′ = 2x− 4y

=⇒ y′ =2x− 4y3y2 + 4x


. . . . . .

Another Example

Example



3y2y′ + 4(1 · y+ x · y′) = 2x


3y2y′ + 4xy′ = 2x− 4y

(3y2 + 4x)y′ = 2x− 4y

=⇒ y′ =2x− 4y3y2 + 4x


. . . . . .

Yet Another Example

Example

Find y′ if y5 + x2y3 = 1+ y sin(x2).

SolutionDifferentiating implicitly:

5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)

Collect all terms with y′ on one side and all terms without y′ on theother:

5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)

Now factor and divide:

y′ =2xy(cos x2 − y2)

5y4 + 3x2y2 − sin x2


. . . . . .

Yet Another Example

Example

Find y′ if y5 + x2y3 = 1+ y sin(x2).

SolutionDifferentiating implicitly:

5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)

Collect all terms with y′ on one side and all terms without y′ on theother:

5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)

Now factor and divide:

y′ =2xy(cos x2 − y2)

5y4 + 3x2y2 − sin x2


. . . . . .

Finding tangent lines with implicit differentitiation.

.

Example

Find the equation of the linetangent to the curve

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

at the point (3,−6).

.

.

Solution

Differentiate: 2ydydx

= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

dydx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

4.

Thus the equation of the tangent line is y+ 6 = −114(x− 3).


. . . . . .


.

Example


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


.

.

Solution


= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

dydx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

4.



. . . . . .


.

Example


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


.

.

Solution


= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

dydx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

4.



. . . . . .

Recall: Line equation forms

I slope-intercept formy = mx+ b

where the slope is m and (0,b) is on the line.I point-slope form

y− y0 = m(x− x0)

where the slope is m and (x0, y0) is on the line.


. . . . . .

Horizontal Tangent Lines

Example

Find the horizontal tangent lines to the same curve: y2 = x3 + x2

SolutionWe have to solve these two equations:

..

y2 = x3 + x2

[(x, y) is on the curve].1.

3x2 + 2x2y

= 0

[tangent lineis horizontal]

.2


. . . . . .

Horizontal Tangent Lines

Example

Find the horizontal tangent lines to the same curve: y2 = x3 + x2

SolutionWe have to solve these two equations:

..

y2 = x3 + x2

[(x, y) is on the curve].1.

3x2 + 2x2y

= 0

[tangent lineis horizontal]

.2


. . . . . .

Solution, continued

I Solving the second equation gives

3x2 + 2x2y

= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0

(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.

I Substituting x = 0 into the first equation gives

y2 = 03 + 02 = 0 =⇒ y = 0

which we’ve disallowed. So no horizontal tangents down that road.I Substituting x = −2/3 into the first equation gives

y2 =

(−23

)3+

(−23

)2=

427

=⇒ y = ± 23√3,

so there are two horizontal tangents.


. . . . . .

Solution, continued


3x2 + 2x2y

= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0

(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.

I Substituting x = 0 into the first equation gives

y2 = 03 + 02 = 0 =⇒ y = 0


y2 =

(−23

)3+

(−23

)2=

427

=⇒ y = ± 23√3,



. . . . . .

Solution, continued


3x2 + 2x2y

= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0

(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.I Substituting x = 0 into the first equation gives

y2 = 03 + 02 = 0 =⇒ y = 0

which we’ve disallowed. So no horizontal tangents down that road.

I Substituting x = −2/3 into the first equation gives

y2 =

(−23

)3+

(−23

)2=

427

=⇒ y = ± 23√3,



. . . . . .

Solution, continued


3x2 + 2x2y

= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0

(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.I Substituting x = 0 into the first equation gives

y2 = 03 + 02 = 0 =⇒ y = 0


y2 =

(−23

)3+

(−23

)2=

427

=⇒ y = ± 23√3,

so there are two horizontal tangents.V63.0121.041, Calculus I (NYU) Section 2.6 Implicit Differentiation October 13, 2010 16 / 34

. . . . . .

Horizontal Tangents

..

.(−2

3 ,2

3√3

).

.(−2

3 ,−2

3√3

)

.

.node

..(−1,0)


. . . . . .

Horizontal Tangents

..

.(−2

3 ,2

3√3

).

.(−2

3 ,−2

3√3

) .

.node

..(−1,0)


. . . . . .

Example

Find the vertical tangent lines to the same curve: y2 = x3 + x2

Solution

I Tangent lines are vertical whendxdy

= 0.

I Differentiating x implicitly as a function of y gives

2y = 3x2dxdy

+ 2xdxdy

, sodxdy

=2y

3x2 + 2x(notice this is the

reciprocal of dy/dx).I We must solve

.

.y2 = x3 + x2

[(x, y) is onthe curve]

.1.

2y3x2 + 2x

= 0[tangent lineis vertical]

.2


. . . . . .

Example


Solution


= 0.


2y = 3x2dxdy

+ 2xdxdy

, sodxdy

=2y



.

.y2 = x3 + x2


.1.

2y3x2 + 2x


.2


. . . . . .

Example


Solution


= 0.


2y = 3x2dxdy

+ 2xdxdy

, sodxdy

=2y


reciprocal of dy/dx).

I We must solve

.

.y2 = x3 + x2


.1.

2y3x2 + 2x


.2


. . . . . .

Example


Solution


= 0.


2y = 3x2dxdy

+ 2xdxdy

, sodxdy

=2y



.

.y2 = x3 + x2


.1.

2y3x2 + 2x


.2V63.0121.041, Calculus I (NYU) Section 2.6 Implicit Differentiation October 13, 2010 18 / 34

. . . . . .

Solution, continued


2y3x2 + 2x

= 0 =⇒ 2y = 0 =⇒ y = 0

(as long as 3x2 + 2x ̸= 0).

I Substituting y = 0 into the first equation gives

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

So x = 0 or x = −1.I x = 0 is not allowed by the first equation, but

dxdy

∣∣∣∣(−1,0)

= 0,

so here is a vertical tangent.


. . . . . .

Solution, continued


2y3x2 + 2x

= 0 =⇒ 2y = 0 =⇒ y = 0

(as long as 3x2 + 2x ̸= 0).I Substituting y = 0 into the first equation gives

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

So x = 0 or x = −1.

I x = 0 is not allowed by the first equation, but

dxdy

∣∣∣∣(−1,0)

= 0,

so here is a vertical tangent.


. . . . . .

Solution, continued


2y3x2 + 2x

= 0 =⇒ 2y = 0 =⇒ y = 0

(as long as 3x2 + 2x ̸= 0).I Substituting y = 0 into the first equation gives

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

So x = 0 or x = −1.I x = 0 is not allowed by the first equation, but

dxdy

∣∣∣∣(−1,0)

= 0,

so here is a vertical tangent.V63.0121.041, Calculus I (NYU) Section 2.6 Implicit Differentiation October 13, 2010 19 / 34

. . . . . .

Tangents

..

.(−2

3 ,2

3√3

).

.(−2

3 ,−2

3√3

) .

.node

..(−1,0)


. . . . . .

Examples

Example

Show that the families of curves

xy = c x2 − y2 = k

are orthogonal, that is, they intersect at right angles.

Solution

I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx

I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy

I The product is −1, so the tangent lines are perpendicularwherever they intersect.


. . . . . .

Orthogonal Families of Curves

xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .


xy = cx2 − y2 = k . .x

.y

.xy=1

.xy=2

.xy=3

.xy=−1

.xy=−2

.xy=−3

.x2−

y2=1

.x2−

y2=2

.x2−

y2=3

.x2 − y2 = −1

.x2 − y2 = −2

.x2 − y2 = −3


. . . . . .

Examples

Example


xy = c x2 − y2 = k


Solution





. . . . . .

Examples

Example


xy = c x2 − y2 = k


Solution





. . . . . .

Examples

Example


xy = c x2 − y2 = k


Solution





. . . . . .

Music Selection

“The Curse of Curves” by Cute is What We Aim ForV63.0121.041, Calculus I (NYU) Section 2.6 Implicit Differentiation October 13, 2010 24 / 34

. . . . . .

Ideal gases

The ideal gas law relatestemperature, pressure, andvolume of a gas:

PV = nRT

(R is a constant, n is theamount of gas in moles)

.

.Image credit: Scott Beale / Laughing SquidV63.0121.041, Calculus I (NYU) Section 2.6 Implicit Differentiation October 13, 2010 25 / 34

http://www.laughingsquid.com/

. . . . . .

Compressibility

DefinitionThe isothermic compressibility of a fluid is defined by

β = −dVdP

1V

with temperature held constant.

Approximately we have

∆V∆P

≈ dVdP

= −βV =⇒ ∆VV

≈ −β∆P

The smaller the β, the “harder” the fluid.


. . . . . .

Compressibility

DefinitionThe isothermic compressibility of a fluid is defined by

β = −dVdP

1V

with temperature held constant.

Approximately we have

∆V∆P

≈ dVdP

= −βV =⇒ ∆VV

≈ −β∆P

The smaller the β, the “harder” the fluid.


. . . . . .

Compressibility of an ideal gas

Example

Find the isothermic compressibility of an ideal gas.

SolutionIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then

dPdP

· V+ PdVdP

= 0 =⇒ dVdP

= −VP

Soβ = −1

V· dVdP

=1P

Compressibility and pressure are inversely related.


. . . . . .

Compressibility of an ideal gas

Example

Find the isothermic compressibility of an ideal gas.

SolutionIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then

dPdP

· V+ PdVdP

= 0 =⇒ dVdP

= −VP

Soβ = −1

V· dVdP

=1P

Compressibility and pressure are inversely related.


. . . . . .

Nonideal gassesNot that there's anything wrong with that

Example

The van der Waals equationmakes fewer simplifications:(

P+ an2

V2

)(V− nb) = nRT,

where P is the pressure, V thevolume, T the temperature, nthe number of moles of the gas,R a constant, a is a measure ofattraction between particles ofthe gas, and b a measure ofparticle size.

...Oxygen..H

..H

..Oxygen

..H

..H

..Oxygen ..H

..H

.

.

.Hydrogen bonds


. . . . . .

Nonideal gassesNot that there's anything wrong with that

Example

The van der Waals equationmakes fewer simplifications:(

P+ an2

V2

)(V− nb) = nRT,

where P is the pressure, V thevolume, T the temperature, nthe number of moles of the gas,R a constant, a is a measure ofattraction between particles ofthe gas, and b a measure ofparticle size.

.

..Wikimedia Commons


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

. . . . . .

Compressibility of a van der Waals gas

Differentiating the van der Waals equation by treating V as a functionof P gives (

P+an2

V2

)dVdP

+ (V− bn)(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Question

I What if a = b = 0?I Without taking the derivative, what is the sign of

dβdb

?

I Without taking the derivative, what is the sign ofdβda

?


. . . . . .



P+an2

V2

)dVdP

+ (V− bn)(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Question


dβdb

?


?


. . . . . .



P+an2

V2

)dVdP

+ (V− bn)(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Question


dβdb

?


?


. . . . . .



P+an2

V2

)dVdP

+ (V− bn)(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Question

I What if a = b = 0?

I Without taking the derivative, what is the sign ofdβdb

?


?


. . . . . .



P+an2

V2

)dVdP

+ (V− bn)(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Question


dβdb

?


?


. . . . . .



P+an2

V2

)dVdP

+ (V− bn)(1− 2an2

V3dVdP

)= 0,

so

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Question


dβdb

?


?


. . . . . .

Nasty derivatives

I

dβdb

= −(2abn3 − an2V+ PV3)(nV2)− (nbV2 − V3)(2an3)(2abn3 − an2V+ PV3)2

= −nV3

(an2 + PV2

)(PV3 + an2(2bn− V)

)2 < 0

Idβda

=n2(bn− V)(2bn− V)V2(PV3 + an2(2bn− V)

)2 > 0

(as long as V > 2nb, and it’s probably true that V ≫ 2nb).


. . . . . .

Outline





. . . . . .

Using implicit differentiation to find derivatives

Example

Finddydx

if y =√x.

SolutionIf y =

√x, then

y2 = x,

so2y

dydx

= 1 =⇒ dydx

=12y

=1

2√x.


. . . . . .

Using implicit differentiation to find derivatives

Example

Finddydx

if y =√x.

SolutionIf y =

√x, then

y2 = x,

so2y

dydx

= 1 =⇒ dydx

=12y

=1

2√x.


. . . . . .


TheoremIf y = xp/q, where p and q are integers, then y′ =

pqxp/q−1.

Proof.First, raise both sides to the qth power:

y = xp/q =⇒ yq = xp

Now, differentiate implicitly:

qyq−1dydx

= pxp−1 =⇒ dydx

=pq· x

p−1

yq−1

Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so

xp−1

yq−1 =xp−1

xp−p/q = xp−1−(p−p/q) = xp/q−1


. . . . . .



pqxp/q−1.




qyq−1dydx

= pxp−1 =⇒ dydx

=pq· x

p−1

yq−1


xp−1

yq−1 =xp−1



. . . . . .



pqxp/q−1.




qyq−1dydx

= pxp−1 =⇒ dydx

=pq· x

p−1

yq−1


xp−1

yq−1 =xp−1



. . . . . .



pqxp/q−1.




qyq−1dydx

= pxp−1 =⇒ dydx

=pq· x

p−1

yq−1


xp−1

yq−1 =xp−1



. . . . . .

Summary

I Implicit Differentiation allows us to pretend that a relationdescribes a function, since it does, locally, “almost everywhere.”

I The Power Rule was established for powers which are rationalnumbers.
