Lesson 11: Implicit Differentiation (Section 21 handout)

Section 2.6Implicit Differentiation

V63.0121.021, Calculus I

New York University

October 11, 2010

Announcements

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

I 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

V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 2 / 34

Objectives

I Use implicit differentation tofind the derivative of afunction defined implicitly.


Notes

Notes

Notes

1

Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010

Outline

The big idea, by example

ExamplesBasic ExamplesVertical and Horizontal TangentsOrthogonal TrajectoriesChemistry

The power rule for rational powers


Motivating Example

Problem

Find the slope of the linewhich is tangent to the curve

x2 + y 2 = 1

at the point (3/5,−4/5).

x

y

Solution (Explicit)

I Isolate: y 2 = 1− x2 =⇒ y = −√

1− x2. (Why the −?)

I Differentiate:dy

dx= − −2x

2√

1− x2=

x√1− x2

I Evaluate:dy

dx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/5

4/5=

3

4.


Motivating Example, another way

We know that x2 + y 2 = 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) = − x

f (x)


Notes

Notes

Notes

2


Yes, we can!

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

I “Near” most points on thecurve x2 + y 2 = 1, the curveresembles the graph of afunction.

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

I The chain rule then appliesfor this local choice.

x

y

looks like a function

looks like a function

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


Motivating Example, again, with Leibniz notation

Problem

Find the slope of the line which is tangent to the curve x2 + y 2 = 1 at thepoint (3/5,−4/5).

Solution

I Differentiate: 2x + 2ydy

dx= 0

Remember y is assumed to be a function of x!

I Isolate:dy

dx= −x

y.

I Evaluate:dy

dx

∣∣∣∣( 3

5,− 4

5 )=

3/5

4/5=

3

4.


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 be assumedto be a function of x

I we may differentiate the relationas is

I Solving fordy

dxdoes give the

slope of the tangent line to thecurve at a point on the curve.

x

y


Notes

Notes

Notes

3


Outline





Another Example

Example

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

Solution

Implicitly differentiating, we have

3y 2y ′ + 4(1 · y + x · y ′) = 2x

Solving for y ′ gives

3y 2y ′ + 4xy ′ = 2x − 4y

(3y 2 + 4x)y ′ = 2x − 4y

=⇒ y ′ =2x − 4y

3y 2 + 4x


Yet Another Example

Example

Find y ′ if y 5 + x2y 3 = 1 + y sin(x2).

Solution

Differentiating implicitly:

5y 4y ′ + (2x)y 3 + x2(3y 2y ′) = y ′ sin(x2) + y cos(x2)(2x)

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

5y 4y ′ + 3x2y 2y ′ − sin(x2)y ′ = −2xy 3 + 2xy cos(x2)

Now factor and divide:

y ′ =2xy(cos x2 − y 2)

5y 4 + 3x2y 2 − sin x2


Notes

Notes

Notes

4


Finding tangent lines with implicit differentitiation

Example

Find the equation of the line tangentto the curve

y 2 = x2(x + 1) = x3 + x2

at the point (3,−6).

Solution

Differentiate: 2ydy

dx= 3x2 + 2x , so

dy

dx=

3x2 + 2x

2y, and

dy

dx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

4.

Thus the equation of the tangent line is y + 6 = −11

4(x − 3).


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 formy − 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: y 2 = x3 + x2

Solution

We have to solve these two equations:

y 2 = x3 + x2

[(x , y) is on the curve]13x2 + 2x

2y= 0

[tangent lineis horizontal]

2


Notes

Notes

Notes

5


Solution, continued

I Solving the second equation gives

3x2 + 2x

2y= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0

(as long as y 6= 0). So x = 0 or 3x + 2 = 0.

I Substituting x = 0 into the first equation gives

y 2 = 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

y 2 =

(−2

3

)3

+

(−2

3

)2

=4

27=⇒ y = ± 2

3√

3,

so there are two horizontal tangents.


Tangents

(−2

3 ,2

3√

3

)

(−2

3 ,−2

3√

3

)node

(−1, 0)


Example

Find the vertical tangent lines to the same curve: y 2 = x3 + x2

Solution

I Tangent lines are vertical whendx

dy= 0.

I Differentiating x implicitly as a function of y gives

2y = 3x2 dx

dy+ 2x

dx

dy, so

dx

dy=

2y

3x2 + 2x(notice this is the reciprocal

of dy/dx).

I We must solve

y 2 = x3 + x2

[(x , y) is onthe curve]

12y

3x2 + 2x= 0

[tangent lineis vertical]

2


Notes

Notes

Notes

6


Solution, continued

I Solving the second equation gives

2y

3x2 + 2x= 0 =⇒ 2y = 0 =⇒ y = 0

(as long as 3x2 + 2x 6= 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

dx

dy

∣∣∣∣(−1,0)

= 0,

so here is a vertical tangent.


Examples

Example

Show that the families of curves

xy = c x2 − y 2 = k

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

Solution

In the first curve,

y + xy ′ = 0 =⇒ y ′ = −y

x

In the second curve,

2x − 2yy ′ = 0 = =⇒ y ′ =x

y

The product is −1, so the tangent lines are perpendicular wherever theyintersect.


Orthogonal Families of Curves

xy = cx2 − y 2 = 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 − y 2 = −1x2 − y 2 = −2x2 − y 2 = −3


Notes

Notes

Notes

7


Examples

Example

Show that the families of curves

xy = c x2 − y 2 = k

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

Solution

In the first curve,

y + xy ′ = 0 =⇒ y ′ = −y

x

In the second curve,

2x − 2yy ′ = 0 = =⇒ y ′ =x

y

The product is −1, so the tangent lines are perpendicular wherever theyintersect.


Ideal gases

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

PV = nRT

(R is a constant, n is the amountof gas in moles)

Image credit: Scott Beale / Laughing SquidV63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 25 / 34

Compressibility

Definition

The isothermic compressibility of a fluid is defined by

β = −dV

dP

1

V

with temperature held constant.

Approximately we have

∆V

∆P≈ dV

dP= −βV =⇒ ∆V

V≈ −β∆P

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


Notes

Notes

Notes

8


http://www.laughingsquid.com/

Compressibility of an ideal gas

Example

Find the isothermic compressibility of an ideal gas.

Solution

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

dP

dP· V + P

dV

dP= 0 =⇒ dV

dP= −V

P

So

β = − 1

V· dV

dP=

1

P

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

V 2

)(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.

OxygenH

H

Oxygen

H

H

Oxygen H

H

Hydrogen bonds

Image credit: Wikimedia Commons


Compressibility of a van der Waals gas

Differentiating the van der Waals equation by treating V as a function ofP gives (

P +an2

V 2

)dV

dP+ (V − bn)

(1− 2an2

V 3

dV

dP

)= 0,

so

β = − 1

V

dV

dP=

V 2(V − nb)

2abn3 − an2V + PV 3

Question

I What if a = b = 0?

I Without taking the derivative, what is the sign ofdβ

db?

I Without taking the derivative, what is the sign ofdβ

da?


Notes

Notes

Notes

9


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

Nasty derivatives

I

dβ

db= −(2abn3 − an2V + PV 3)(nV 2)− (nbV 2 − V 3)(2an3)

(2abn3 − an2V + PV 3)2

= −nV 3

(an2 + PV 2

)(PV 3 + an2(2bn − V ))2

< 0

I

dβ

da=

n2(bn − V )(2bn − V )V 2

(PV 3 + 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

Finddy

dxif y =

√x .

Solution

If y =√

x, theny 2 = x ,

so

2ydy

dx= 1 =⇒ dy

dx=

1

2y=

1

2√

x.


Notes

Notes

Notes

10



Theorem

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

qxp/q−1.

Proof.

First, raise both sides to the qth power:

y = xp/q =⇒ yq = xp

Now, differentiate implicitly:

qyq−1 dy

dx= pxp−1 =⇒ dy

dx=

p

q· xp−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


Summary

I Implicit Differentiation allows us to pretend that a relation describes afunction, since it does, locally, “almost everywhere.”

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


Notes

Notes

Notes

11


Documents

Lesson 11: Implicit Differentiation (Section 21 handout)