Lesson 11: Implicit Differentiation (Section 21 slides)

Section 2.6
Implicit Differentiation

V63.0121.021, Calculus I

New York University

October 12, 2010

Announcements

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

. . . . . .

Announcements

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

. . . . . .

V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 2 / 34

Objectives

Use implicit differentation
to find the derivative of a
function defined implicitly.

. . . . . .


Outline

The big idea, by example

Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry

. . . . . .


Motivating Example
y
.
Problem
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
y
.
Problem
curve . x
.
x2 + y2 = 1
.

. . . . . .


Motivating Example
y
.
Problem
curve . x
.
x2 + y2 = 1
.

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

. . . . . .


Motivating Example
y
.
Problem
curve . x
.
x2 + y2 = 1
.

Solution (Explicit)
√
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2

. . . . . .


Motivating Example
y
.
Problem
curve . x
.
x2 + y2 = 1
.

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

. . . . . .


suppose it did.

x2 + (f(x))2 = 1

We could differentiate this equation to get

2x + 2f(x) · f′ (x) = 0

. . . . . .


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!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.

. x
.

.

. . . . . .


Yes, we can!

y
.
of a function.

. x
.

.

l
.ooks like a function

. . . . . .


Yes, we can!

y
.
curve x2 + y2 = 1, the .
of a function.

. x
.

. . . . . .


Yes, we can!

y
.
curve x2 + y2 = 1, the .
of a function. l
. x
.

. . . . . .


Yes, we can!

y
.
of a function.

. . x
.

. . . . . .


Yes, we can!

y
.
of a function.

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

. . . . . .


Yes, we can!

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

.

l

. . . . . .


Yes, we can!

y
.
of a function.
So f(x) is defined “locally”,
. x
.
almost 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
the point (3/5, −4/5).

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

. . . . . .



Problem
the point (3/5, −4/5).

Solution
dy
Differentiate: 2x + 2y
=0
dx
Remember y is assumed to be a function of x!

. . . . . .



Problem
the point (3/5, −4/5).

Solution
dy
Differentiate: 2x + 2y
=0
dx
dy x
Isolate: =− .
dx y

. . . . . .



Problem
the point (3/5, −4/5).

Solution
dy
Differentiate: 2x + 2y =0
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 y
.
most places” y can be
.
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.

. . . . . .


Outline


Examples
Basic Examples
Chemistry

. . . . . .


Another Example

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

. . . . . .


Another Example

Example

Solution
Implicitly differentiating, we have

3y2 y′ + 4(1 · y + x · y′ ) = 2x

. . . . . .


Another Example

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

. . . . . .


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

. . . . . .


Yet Another Example
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
. . . . . .


Finding tangent lines with implicit differentitiation
.

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

at the point (3, −6). .

. . . . . . .


.

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


Solution
dy dy 3x2 + 2x
Differentiate: 2y = 3x2 + 2x, so = , and
dx dx 2y

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

. . . . . . .


.

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


Solution
dy dy 3x2 + 2x
Differentiate: 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
. . . . . . .


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

Solution
We have to solve these two equations:

.
. 3x2 + 2x
2 3 2 = 0
1 y = x. + x
. [(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 3 2 2 4 2
y = −
2
+ − = =⇒ y = ± √ ,
3 3 27 3 3
so there are two horizontal tangents.
. . . . . .


Horizontal Tangents

( )
. −2,
3
2
√
3 3
.
.
.
( )
. −2, −
3
2
√
3 3

. . . . . .


Horizontal Tangents

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

. . . . . .


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 = x3 + x2 =0
3x2 + 2x
1
. [(x, y). is on
the curve]
2
. [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
2
√
3 3
.
. −1, 0) .
( .
.
( )
. −2, −
3
2
√ n
. ode
3 3

. . . . . .


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
=
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
.

.xy = −1
= −2
.xy =
.xy

3
−
. . . . . .


y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

.xy = −1
= −2
x

.xy =
.xy

3
−
. . . . . .


y
.

.xy
.xy
=
3
=
.xy

2
=
. 2 − y2 = 2
. 2 − y2 = 1

1
xy = c
x2 − y2 = k . x
.

.xy = −1
= −2
x
x

.xy =
.xy

3
−
. . . . . .


y
.

.xy
.xy
=
3
=
.xy

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

1
xy = c
x2 − y2 = k . x
.

.xy = −1
x2

= −2
x
x

.xy =
.xy

3
−
. . . . . .


y
.

.xy
.xy
=
3
=
.xy

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

1
xy = c
x2 − y2 = k . x
.

.xy = −1
x2

= −2
x
x
. 2 − y2 = −1

.xy =
x

.xy

3
−
. . . . . .


y
.

.xy
.xy
=
3
=
.xy

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

1
xy = c
x2 − y2 = k . x
.

.xy = −1
x2

= −2
x
x
. 2 − y2 = −1

.xy =
x

.xy

3
. 2 − y2 = −2

−
x

. . . . . .


y
.

.xy
.xy
=
3
=
.xy

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

1
xy = c
x2 − y2 = k . x
.

.xy = −1
x2

= −2
x
x
. 2 − y2 = −1

.xy =
x

.xy

3
. 2 − y2 = −2

−
x2
. − y2 = −3
x

. . . . . .


Examples

Example

xy = c x2 − y2 = k


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

. . . . . .


Examples

Example

xy = c x2 − y2 = k


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

. . . . . .


Examples

Example

xy = c x2 − y2 = k


Solution
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.

. . . . . .


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)

.
Ima
.
. . . . . .


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
≈ = −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.

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 H
..
makes fewer simplifications:
( ) O .
. xygen . .
H
n2 .
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 measure of O .
. xygen . .
H
attraction between particles of
the gas, and b a measure of H
..
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.
. . ikimedia Commons
W
. . . . . .


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

. . . . . .


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

. . . . . .


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=− =

Question

. . . . . .


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=− =

Question

What if a = b = 0?

. . . . . .


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=− =

Question

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

. . . . . .


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=− =

Question

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
( )
nV3 an2 + PV2
= −( )2 < 0
PV3 + an2 (2bn − V)

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

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

. . . . . .


Outline


Examples
Basic Examples
Chemistry

. . . . . .


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

. . . . . .


The power rule for rational powers
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

xp−1 xp−1
= p−p/q = xp−1−(p−p/q) = xp/q−1
. . . . . .

V63.0121.021, Calculus I
yq−1
(NYU)
x Section 2.6 Implicit Differentiation October 12, 2010 33 / 34

Summary

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 (Section 21 slides)

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