Lesson 11: Implicit Differentiation (slides)

Sec on 2.6
Implicit Diﬀeren a on
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

February 28, 2011

.

Music Selection

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

Announcements

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

Objectives

Use implicit differenta on
to find the deriva ve of a
func on defined
implicitly.

Outline
The big idea, by example

Examples
Basic Examples
Ver cal and Horizontal Tangents
Orthogonal Trajectories
Chemistry

The power rule for ra onal powers

Motivating Example

Problem y
Find the slope of the line
which is tangent to the
curve . x
x2 + y2 = 1

at the point (3/5, −4/5).

Motivating Example, Solution
Solu on (Explicit)

Isolate: √ y
y = 1 − x =⇒ y = − 1 − x2 .
2 2

(Why the −?)
. x

Solu on (Explicit)

Isolate: √ y
y = 1 − x =⇒ y = − 1 − x2 .
2 2

(Why the −?)
Diﬀeren ate: . x
dy −2x x
=− √ =√
dx 2 1 − x2 1 − x2

Solu on (Explicit)

Isolate: √ y
y = 1 − x =⇒ y = − 1 − x2 .
2 2

(Why the −?)
Diﬀeren ate: . x
dy −2x x
=− √ =√
dx 2 1 − x2 1 − x2
Evaluate:
dy 3/5 3/5 3
=√ = = .
dx x=3/5 1 − (3/5)2 4/5 4

Motivating 2Example, another way
2
We know that x + y = 1 does not deﬁne y as a func on of x, but
suppose it did.
Suppose we had y = f(x), so that
x2 + (f(x))2 = 1

2
suppose it did.
x2 + (f(x))2 = 1
We could diﬀeren ate this equa on to get
2x + 2f(x) · f′ (x) = 0

2
suppose it did.
x2 + (f(x))2 = 1
We could diﬀeren ate this equa on to get
2x + 2f(x) · f′ (x) = 0
We could then solve to get
x
f′ (x) = −
f(x)

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

. x

Yes, we can!
y

. x

looks like a func on

Yes, we can!
y
. x

Yes, we can!
y

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

Yes, we can!
y
So f(x) is deﬁned “locally”,
almost everywhere and is . x
diﬀeren able


Yes, we can!
y
So f(x) is deﬁned “locally”,
almost everywhere and is . x
diﬀeren able
The chain rule then applies for
this local choice.

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
the point (3/5, −4/5).

Solu on
dy
Diﬀeren ate: 2x + 2y =0
dx

Problem
the point (3/5, −4/5).

Solu on
dy
Diﬀeren ate: 2x + 2y = 0
dx
Remember y is assumed to be a func on of x!

Problem
the point (3/5, −4/5).

Solu on
dy
dx
dy x
Isolate: =− .
dx y

Problem
the point (3/5, −4/5).

Solu on
dy
dx
dy x dy 3/5 3
Isolate: = − . Then evaluate: = = .
dx y dx ( 3 ,− 4 ) 4/5 4
5 5

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

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

Another Example
Example

Solu on
Implicitly diﬀeren a ng, we have

3y2 y′ + 4(1 · y + x · y′ ) = 2x

Another Example
Example

Solu on
Implicitly diﬀeren a ng, we have

3y2 y′ + 4(1 · y + x · y′ ) = 2x

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

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

Yet Another Example
Example

Solu on
Diﬀeren a ng implicitly:

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

Yet Another Example
Example

Solu on
Diﬀeren a ng 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 )

Yet Another Example
Example

Solu on
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 )

2xy(cos x2 − y2 )
′
Now factor and divide: y = 4
5y + 3x2 y2 − sin x2

Finding tangents with implicit diﬀerentitiation

Example
Find the equa on of the line
tangent to the curve
.
y2 = x2 (x + 1) = x3 + x2

at the point (3, −6).

Solution
Solu on
dy 2 dy 3x2 + 2x
Diﬀeren ate: 2y = 3x + 2x, so = , and
dx dx 2y

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

Solution
Solu on
dy 2 dy 3x2 + 2x
Diﬀeren ate: 2y = 3x + 2x, so = , and
dx dx 2y

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

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

Recall: 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.

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

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

Solu on
We have to solve these two equa ons:
y2 = x3 + x2 [(x, y) is on the curve]
3x2 + 2x
= 0 [tangent line is horizontal]
2y

Solution, continued
Solving the second equa on 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
(as long as y ̸= 0). So x = 0 or 3x + 2 = 0.
Subs tu ng x = 0 into the ﬁrst equa on gives

y2 = 03 + 02 = 0 =⇒ y = 0

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

Solution, continued
Subs tu ng x = −2/3 into the ﬁrst equa on gives
( )3 ( )2
2 2 4
y = −
2
+ − =
3 3 27
√
4 2
=⇒ y = ± =± √ ,
27 3 3
so there are two horizontal tangents.

Horizontal Tangents

( )
− 3 , 3√3
2 2

.
( )
− 2 , − 3√3
3
2

Horizontal Tangents

( )
− 3 , 3√3
2 2

.
( )
− 2 , − 3√3
3
2
node

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

Example

Solu on
dx
Tangent lines are ver cal when = 0.
dy

Example

Solu on
dx
dy
Diﬀeren a ng x implicitly as a func on of y gives
dx dx dx 2y
2y = 3x2 + 2x , so = 2 (no ce this is the
dy dy dy 3x + 2x
reciprocal of dy/dx).

Example

Solu on
dx
dy
Diﬀeren a ng x implicitly as a func on of y gives
dx dx dx 2y
2y = 3x2 + 2x , so = 2 (no ce this is the
dy dy dy 3x + 2x
reciprocal of dy/dx).
We must solve y2 = x3 + x2 [(x, y) is on the curve] and
2y
= 0 [tangent line is ver cal]
3x2 + 2x

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).
Subs tu ng y = 0 into the ﬁrst equa on 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).
Subs tu ng y = 0 into the ﬁrst equa on gives

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

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

Tangents

( )
− 3 , 3√3
2 2

(−1, 0) .
( )
− 3 , − 3√3 node
2 2

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
=
xy = c

1
x2 − y2 = k . x

y

xy y =
x
= 1
2
xy = c
x2 − y2 = k . x

y

xy = 1
xy y =
= 2
x

3
xy = c
x2 − y2 = k . x

y

xy = 1
xy y =
= 2
x

3
xy = c
x2 − y2 = k . x

1
−
=
xy

y

xy = 1
xy y =
= 2
x

3
xy = c
x2 − y2 = k . x

− 1
= −
2
xy y =
x

y

xy = 1
xy y =
= 2
x

3
xy = c
x2 − y2 = k . x

= − 1
xy = −
− 2
xy y =

3
x

y

xy = 1
xy y =
= 2
x

3
x2 − y2 = 1
xy = c
x2 − y2 = k . x

= − 1
xy = −
− 2
xy y =

3
x

y

xy = 1
xy y =
= 2
x

3
x2 − y2 = 2
x −y =1
xy = c
x2 − y2 = k . x

2

= − 1
xy = −
− 2
xy y =

3
2

x

xy = c
x2 − y2 = k

2 2
x2 − y2 = 3
x2 − y2 = 2
x −y =1
x
.
y

xy y = x
xy = −
= − 1 xy y =
− 2
xy = 1
3 = 2
3
x

y

xy = 1
xy y =
= 2
x

3
x2 − y2 = 3
x2 − y2 = 2
x −y =1
xy = c
x2 − y2 = k . x

2

= − 1
xy = −
− 2
x2 − y2 = −1

xy y =

3
2

x

y

xy = 1
xy y =
= 2
x

3
x2 − y2 = 3
x2 − y2 = 2
x −y =1
xy = c
x2 − y2 = k . x

2

= − 1
xy = −
− 2
x2 − y2 = −1

xy y =

3
2
x2 − y2 = −2

x

y

xy = 1
xy y =
= 2
x

3
x2 − y2 = 3
x2 − y2 = 2
x −y =1
xy = c
x2 − y2 = k . x

2

= − 1
xy = −
− 2
x2 − y2 = −1

xy y =

3
2
x2 − y2 = −2
x2 − y2 = −3

x

Examples
Example

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

Examples
Example

Solu on
y
x
′ ′ x
In the second curve, 2x − 2yy = 0 =⇒ y =
y

Examples
Example

Solu on
y
x
′ ′ x
In the second curve, 2x − 2yy = 0 =⇒ y =
y
The product is −1, so the tangent lines are perpendicular
wherever they intersect.

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: Sco Beale / Laughing Squid

Compressibility
Defini on
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V

Compressibility
Defini on
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V

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

Compressibility of an ideal gas
Example
Find the isothermic compressibility of an ideal gas.

Compressibility of an ideal gas
Example
Find the isothermic compressibility of an ideal gas.

Solu on
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
1 dV 1
So β = − · = . Compressibility and pressure are inversely
V dP P
related.

Nonideal gasses
Not that there’s anything wrong with that
Example
H.

The van der Waals equa on makes
fewer simpliﬁca ons:
. .
Oxygen H
( ) H.
n2 .
P + a 2 (V − nb) = nRT, Oxygen Hydrogen bonds
V H.

where a is a measure of a rac on . .
Oxygen H
between par cles of the gas, and b a
measure of par cle size. H.

Nonideal gasses
Not that there’s anything wrong with that
Example
The van der Waals equa on makes
fewer simpliﬁca ons:
( )
n2 .
P + a 2 (V − nb) = nRT,
V
where a is a measure of a rac on
between par cles of the gas, and b a
measure of par cle size.

Compressibility of a van der Waals gas

Diﬀeren a ng the van der Waals equa on by trea ng V as a
func on of P gives
( ) ( )
an2 dV 2an2 dV
P+ 2 + (V − bn) 1 − 3 = 0,
V dP V dP

Compressibility of a van der Waals gas

Diﬀeren a ng the van der Waals equa on by trea ng V as a
func on 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

Nonideal compressibility,
continued
1 dV V2 (V − nb)
β=− =

Ques on

continued
1 dV V2 (V − nb)
β=− =

Ques on

What if a = b = 0?

continued
1 dV V2 (V − nb)
β=− =

Ques on

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

continued
1 dV V2 (V − nb)
β=− =

Ques on

What if a = b = 0?
dβ
db
dβ
da

Nasty derivatives
Answer
We get the old (ideal) compressibility
We have
( )
dβ nV3 an2 + PV2
= −( )2 < 0
db PV3 + an2 (2bn − V)

dβ n2 (bn − V)(2bn − V)V2
We have =( ) > 0 (as long as
da 3
PV + an 2 (2bn − V) 2

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
Solu on
√
If y = x, then
y2 = x,
so
dy dy 1 1
2y = 1 =⇒ = = √ .
dx dx 2y 2 x

The power rule for rational powers
Theorem
p
If y = xp/q , where p and q are integers, then y′ = xp/q−1 .
q

Theorem
p
q

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

y = xp/q =⇒ yq = xp

Theorem
p
q

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

y = xp/q =⇒ yq = xp

Now, diﬀeren ate implicitly:

q−1 dy dy p xp−1
qy = px p−1
=⇒ = ·
dx dx q yq−1

Theorem
p
q

Proof.
Now, diﬀeren ate implicitly:

q−1 dy dy p xp−1
qy = px p−1
=⇒ = ·
dx dx q yq−1

Theorem
p
q

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

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

Summary

Using implicit diﬀeren a on we can treat rela ons which are
not quite func ons like they were func ons.
In par cular, we can ﬁnd the slopes of lines tangent to curves
which are not graphs of func ons.

Lesson 11: Implicit Differentiation (slides)

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