Lesson 11: Implicit Differentiation

Section 2.6
Implicit Differentiation

V63.0121.006/016, Calculus I

February 23, 2010

Announcements
Quiz 2 is February 26, covering §§1.5–2.3
Midterm is March 4, covering §§1.1–2.5
On HW 5, Problem 2.3.46 should be 2.4.46

. .
Image credit: Telstar Logistics
. . . . . .

Announcements on white background

Announcements
Quiz 2 is February 26, covering §§1.5–2.3
Midterm is March 4, covering §§1.1–2.5
On HW 5, Problem 2.3.46 should be 2.4.46

. . . . . .

Outline

The big idea, by example

Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry

The power rule for rational powers

. . . . . .

Motivating Example y
.

Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1

at the point (3/5, −4/5).

. . . . . .

.

Problem
. x
.
2 2
x +y =1

at the point (3/5, −4/5). .

. . . . . .

.

Problem
. x
.
2 2
x +y =1

at the point (3/5, −4/5). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)

. . . . . .

.

Problem
. x
.
2 2
x +y =1

at the point (3/5, −4/5). .
Solution (Explicit)
√
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2

. . . . . .

.

Problem
. x
.
2 2
x +y =1

at the point (3/5, −4/5). .
Solution (Explicit)
√
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2
dy 3 /5 3/5 3
Evaluate: =√ = = .
dx x=3/5 1 − ( 3 /5 )2 4/5 4

. . . . . .

Motivating Example, another way
We know that x2 + y2 = 1 does not deﬁne y as a function of x,
but suppose it did.
Suppose we had y = f(x), so that

x2 + (f(x))2 = 1

. . . . . .

but suppose it did.

x2 + (f(x))2 = 1

We could differentiate this equation to get

2x + 2f(x) · f′ (x) = 0

. . . . . .

but suppose it did.

x2 + (f(x))2 = 1

We could differentiate this equation to get

2x + 2f(x) · f′ (x) = 0

We could then solve to get
x
f′ (x ) = −
f(x)

. . . . . .

Yes, we can!

The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.

. x
.

.

. . . . . .

Yes, we can!

.

. x
.

.

l
.ooks like a function

. . . . . .

Yes, we can!

.
the curve resembles the .

. x
.

. . . . . .

Yes, we can!

.
the curve resembles the .
l
. x
.

. . . . . .

Yes, we can!

.

. . x
.

. . . . . .

Yes, we can!

.

. . x
.
.
does not look like a
function, but that’s
OK—there are only
two points like this

. . . . . .

Yes, we can!

.
So f(x) is deﬁned
“locally”, almost
. x
.
everywhere and is
differentiable

.

l

. . . . . .

Yes, we can!

.
So f(x) is deﬁned
“locally”, almost
. x
.
everywhere and is
differentiable
The chain rule then
.
applies for this local
choice.
l

. . . . . .

Motivating Example, again, with Leibniz notation

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

. . . . . .


Problem
x2 + y2 = 1 at the point (3/5, −4/5).

Solution
dy
Differentiate: 2x + 2y =0
dx

. . . . . .


Problem
x2 + y2 = 1 at the point (3/5, −4/5).

Solution
dy
dx
Remember y is assumed to be a function of x!

. . . . . .


Problem
x2 + y2 = 1 at the point (3/5, −4/5).

Solution
dy
dx
dy x
Isolate: =− .
dx y

. . . . . .


Problem
x2 + y2 = 1 at the point (3/5, −4/5).

Solution
dy
dx
dy x
Isolate: =− .
dx y
dy 3 /5 3
Evaluate: = = .
dx ( 3 ,− 4 ) 4/5 4
5 5

. . . . . .

Summary

If a relation is given between x and y which isn’t a function:
“Most of the time”, i.e., “at
most places” y can be y
.
assumed to be a function of .
x
we may differentiate the . x
.
relation as is
dy
Solving for does give the
dx
slope of the tangent line to
the curve at a point on the
curve.

. . . . . .

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

. . . . . .

Example

Solution
Implicitly differentiating, we have

3y2 y′ + 4(1 · y + x · y′ ) = 2x

. . . . . .

Example

Solution
Implicitly differentiating, we have

3y2 y′ + 4(1 · y + x · y′ ) = 2x

Solving for y′ gives

3y2 y′ + 4xy′ = 2x − 4y
(3y2 + 4x)y′ = 2x − 4y
2x − 4y
=⇒ y′ = 2
3y + 4x

. . . . . .

Example
Find y′ if y5 + x2 y3 = 1 + y sin(x2 ).

. . . . . .

Example
Find y′ if y5 + x2 y3 = 1 + y sin(x2 ).

Solution
Differentiating implicitly:

5y4 y′ + (2x)y3 + x2 (3y2 y′ ) = y′ sin(x2 ) + y cos(x2 )(2x)

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

5y4 y′ + 3x2 y2 y′ − sin(x2 )y′ = −2xy3 + 2xy cos(x2 )

Now factor and divide:
2xy(cos x2 − y2 )
y′ =
5y4 + 3x2 y2 − sin x2

. . . . . .

Example
Find the equation of the line
tangent to the curve
.
y 2 = x 2 (x + 1 ) = x 3 + x 2

at the point (3, −6).
.

. . . . . .

Example
.
y 2 = x 2 (x + 1 ) = x 3 + x 2

.
Solution
Differentiating the expression implicitly with respect to x gives
dy dy 3x2 + 2x
2y = 3x2 + 2x, so = , and
dx dx 2y

dy 3 · 32 + 2 · 3 33 11
= =− =− .
dx (3,−6) 2(−6) 12 4

. . . . . .

Example
.
y 2 = x 2 (x + 1 ) = x 3 + x 2

.
Solution
Differentiating the expression implicitly with respect to x gives
dy dy 3x2 + 2x
2y = 3x2 + 2x, so = , and
dx dx 2y

dy 3 · 32 + 2 · 3 33 11
= =− =− .
dx (3,−6) 2(−6) 12 4

11
Thus the equation of the tangent line is y + 6 = − (x − 3).
4
. . . . . .

Line equation forms

slope-intercept form

y = mx + b

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

y − y0 = m(x − x0 )

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

. . . . . .

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

. . . . . .

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

Solution
We have to solve these two equations:

.
. 2 3 2 3x2 + 2x
y = x +x = 0
1
. [(x, y). is on
the curve]
2
.
2y
[tangent line
is horizontal]

. . . . . .

Solution, continued
Solving the second equation gives

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

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

. . . . . .

Solution, continued

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

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.

. . . . . .

Solution, continued

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

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.
Substituting x = −2/3 into the ﬁrst equation gives
( ) ( )
2 2 3 2 2 4 2
y = − + − = =⇒ y = ± √ ,
3 3 27 3 3
so there are two horizontal tangents.
. . . . . .

Horizontal Tangents

( )
. − 2 , 3√3
3
2

.
.
.
( )
. − 2 , − 3 √3
3
2

. . . . . .

Horizontal Tangents

( )
. − 2 , 3√3
3
2

.
.
.
( )
. − 2 , − 3 √3
3
2 n
. ode

. . . . . .

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

. . . . . .

Example
Solution
dx
Tangent lines are vertical when = 0.
dy

. . . . . .

Example
Solution
dx
dy
Differentiating x implicitly as a function of y gives
dx dx dx 2y
2y = 3x2 + 2x , so = 2 (notice this is the
dy dy dy 3x + 2x
reciprocal of dy/dx).

. . . . . .

Example
Solution
dx
dy
Differentiating x implicitly as a function of y gives
dx dx dx 2y
2y = 3x2 + 2x , so = 2 (notice this is the
dy dy dy 3x + 2x
reciprocal of dy/dx).
We must solve

. .
2y
y2 = x 3 + x2 = 0
1
. [(x, y). is on
the curve]
2
.
3x2 + 2x
[tangent line
is vertical]

. . . . . .

Solution, continued

2y
= 0 =⇒ 2y = 0 =⇒ y = 0
3x2 + 2x

(as long as 3x2 + 2x ̸= 0).

. . . . . .

Solution, continued

2y
= 0 =⇒ 2y = 0 =⇒ y = 0
3x2 + 2x

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

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

So x = 0 or x = −1.

. . . . . .

Solution, continued

2y
= 0 =⇒ 2y = 0 =⇒ y = 0
3x2 + 2x

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

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

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

dx
= 0,
dy (−1,0)

so here is a vertical tangent.
. . . . . .

Tangents

( )
. − 2 , 3√3
3
2

.
. −1 , 0 ) .
( .
.
( )
. − 2 , − 3 √3
3
2 n
. ode

. . . . . .

Examples
Example
Show that the families of curves

xy = c x2 − y2 = k

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

. . . . . .

Orthogonal Families of Curves
y
.

xy = c
x2 − y2 = k . x
.

. . . . . .

y
.

.xy
=
1
xy = c
x2 − y2 = k . x
.

. . . . . .

y
.

.xy
=
.xy

2
=
1
xy = c
x2 − y2 = k . x
.

. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
1
xy = c
x2 − y2 = k . x
.

. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
1
xy = c
x2 − y2 = k . x
.

1
−
=
.xy

. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
1
xy = c
x2 − y2 = k . x
.

1
−
2
=
−
.xy
=
.xy
. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
1
xy = c
x2 − y2 = k . x
.

1
−
− 2
=
−
.xy

3
=
.xy
=
.xy
. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

1
−
− 2
=
x

−
.xy

3
=
.xy
=
.xy
. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 2
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

1
−
− 2
=
x
x

−
.xy

3
=
.xy
=
.xy
. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 3
. − y2 = 2
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

1
−
x2

− 2
=
x
x

−
.xy

3
=
.xy
=
.xy
. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 3
. − y2 = 2
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

1
−
x2 . 2 − y2 = −1

− 2
x

=
x
x

−
.xy

3
=
.xy
=
.xy
. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 3
. − y2 = 2
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

1
−
x2 . 2 − y2 = −1

− 2
x

=
x
x

−
.xy

3
. 2 − y2 = −2
x

=
.xy
=
.xy
. . . . . .

y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 3
. − y2 = 2
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

1
−
x2 . 2 − y2 = −1

− 2
x

=
x
x

−
.xy

3
. 2 − y2 = −2
x2

=
. − y2 = −3

.xy
x

=
.xy
. . . . . .

Examples
Example

xy = c x2 − y2 = k

Solution
In the ﬁrst curve,
y
y + xy′ = 0 =⇒ y′ = −
x

. . . . . .

Examples
Example

xy = c x2 − y2 = k

Solution
In the ﬁrst curve,
y
y + xy′ = 0 =⇒ y′ = −
x
In the second curve,
x
2x − 2yy′ = 0 = =⇒ y′ =
y

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

Music Selection

“The Curse of Curves” by Cute is What We Aim For
. . . . . .

Ideal gases

The ideal gas law relates
temperature, pressure, and
volume of a gas:

PV = nRT

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

.
.
Image credit: Scott Beale / Laughing Squid
. . . . . .

Compressibility

Definition
The isothermic compressibility of a fluid is defined by

dV 1
β=−
dP V
with temperature held constant.

. . . . . .

Compressibility

Definition
The isothermic compressibility of a fluid is defined by

dV 1
β=−
dP V
with temperature held constant.

Approximately we have

∆V dV ∆V
≈ = −β V =⇒ ≈ −β∆P
∆P dP V
The smaller the β , the “harder” the fluid.

. . . . . .

Example
Find the isothermic 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
the word isothermic, and R really is constant), then

dP dV dV V
·V+P = 0 =⇒ =−
dP dP dP P
So
1 dV 1
β=− · =
V dP P
Compressibility and pressure are inversely related.

. . . . . .

Nonideal gasses
Not that there’s anything wrong with that

Example
The van der Waals equation
makes fewer simpliﬁcations: H
..
( ) O .
. xygen . .
n2 H
P + a 2 (V − nb) = nRT, .
V
H
..
where P is the pressure, V the O .
. xygen H
. ydrogen bonds
volume, T the temperature, n H
..

the number of moles of the .
gas, R a constant, a is a O .
. xygen . .
H
measure of attraction
between particles of the gas, H
..
and b a measure of particle
size.
. . . . . .

Nonideal gasses
Not that there’s anything wrong with that

Example
The van der Waals equation
makes fewer simpliﬁcations:
( )
n2
P + a 2 (V − nb) = nRT,
V

where P is the pressure, V the
volume, T the temperature, n
the number of moles of the
gas, R a constant, a is a
measure of attraction
between particles of the gas,
and b a measure of particle
size. .
.
Image credit: Wikimedia Commons
. . . . . .

Let’s ﬁnd the compressibility of a van der Waals gas.
Differentiating the van der Waals equation by treating V as a
function of P gives
( ) ( )
an2 dV 2an2 dV
P+ 2 + (V − bn) 1 − 3 = 0,
V dP V dP

. . . . . .

function of P gives
( ) ( )
an2 dV 2an2 dV
P+ 2 + (V − bn) 1 − 3 = 0,
V dP V dP
so
1 dV V2 (V − nb)
β=− =
V dP 2abn3 − an2 V + PV3

. . . . . .

function of P gives
( ) ( )
an2 dV 2an2 dV
P+ 2 + (V − bn) 1 − 3 = 0,
V dP V dP
so
1 dV V2 (V − nb)
β=− =

What if a = b = 0?

. . . . . .

function of P gives
( ) ( )
an2 dV 2an2 dV
P+ 2 + (V − bn) 1 − 3 = 0,
V dP V dP
so
1 dV V2 (V − nb)
β=− =

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

. . . . . .

function of P gives
( ) ( )
an2 dV 2an2 dV
P+ 2 + (V − bn) 1 − 3 = 0,
V dP V dP
so
1 dV V2 (V − nb)
β=− =

What if a = b = 0?
dβ
db
dβ
da

. . . . . .

Nasty derivatives

dβ (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 )
=−
db (2abn3 − an2 V + PV3 )2
( 2 )
nV3 an + PV2
= −( )2 < 0
PV3 + an2 (2bn − V)

dβ n2 (bn − V)(2bn − V)V2
= ( )2 > 0
da PV3 + an2 (2bn − V)
(as long as V > 2nb, and it’s probably true that V ≫ 2nb).

. . . . . .

Using implicit differentiation to ﬁnd derivatives

Example
dy √
Find if y = x.
dx

. . . . . .

Using implicit differentiation to ﬁnd derivatives

Example
dy √
Find if y = x.
dx
Solution
√
If y = x, then
y2 = x,
so
dy dy 1 1
2y = 1 =⇒ = = √ .
dx dx 2y 2 x

. . . . . .

Theorem
p p/q−1
If y = xp/q , where p and q are integers, then y′ = x .
q

. . . . . .

Theorem
p p/q−1
q
Proof.
First, raise both sides to the qth power:

y = xp/q =⇒ yq = xp

. . . . . .

Theorem
p p/q−1
q
Proof.


Now, differentiate implicitly:

dy dy p xp−1
qyq−1 = pxp−1 =⇒ = · q −1
dx dx q y

. . . . . .

Theorem
p p/q−1
q
Proof.


Now, differentiate implicitly:

dy dy p xp−1
qyq−1 = pxp−1 =⇒ = · q −1
dx dx q y

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

x p −1 xp−1
= p−p/q = xp−1−(p−p/q) = xp/q−1
y q −1 x

. . . . . .

What have we learned today?

Implicit Differentiation allows us to pretend that a relation
describes a function, since it does, locally, “almost
everywhere.”
The Power Rule was established for powers which are
rational numbers.

. . . . . .

Lesson 11: Implicit Differentiation

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Lesson 11: Implicit Differentiation