Lesson 11: Implicit Differentiation

Section 2.6
Implicit Differentiation

V63.0121.034, Calculus I

October 7, 2009

Announcements
Midterm next , covering §§1.1–2.4.

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

Outline

The big idea, by example

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

. . . . . .

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

We could differentiate this equation to get

2x + 2f(x) · f′ (x) = 0

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

. . . . . .

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

. . . . . .

.
So f(x) is deﬁned
.
differentiable
The chain rule then
applies for this local .
choice.
l
.ooks like a function

. . . . . .

.
the curve resembles the .
So f(x) is deﬁned
.
differentiable
The chain rule then
applies for this local
choice.

. . . . . .

.
the curve resembles the .
l
.ooks like a function
So f(x) is deﬁned
.
differentiable
The chain rule then
choice.

. . . . . .

.
So f(x) is deﬁned
“locally” and is . . x
.
differentiable
The chain rule then
choice.

. . . . . .

.
So f(x) is deﬁned
“locally” and is . . x
.
differentiable .
The chain rule then does not look like a
applies for this local function, but that’s
choice. OK—there are only
two points like this

. . . . . .

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

Solution (Implicit, with Leibniz notation)
Differentiate. Remember y is assumed to be a function of x:

dy
2x + 2y = 0,
dx
dy
Isolate :
dx
dy x
=− .
dx y
Evaluate:
dy 3 /5 3
= = .
dx ( 3 ,− 4 ) 4/5 4
5 5

. . . . . .

Summary

If a relation is given between x y
.
and y,
“Most of the time”, i.e., “at
most places” y can be
.
assumed to be a function of
x
.
we may differentiate the
relation as is
dy
Solving for does give the
dx
slope of the tangent line to
the curve at a point on the
curve.

. . . . . .

Mnemonic

Explicit Implicit
y = f(x) F(x, y) = k

. . . . . .

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 11
= =− .
dx (3,−6) 2(−6) 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 11
= =− .
dx (3,−6) 2(−6) 4

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

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

. . . . . .

Example

Solution

. . . . . .

Example

Solution
We solve for dy/dx = 0:

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

. . . . . .

Example

Solution

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

The possible solution x = 0 leads to y = 0, which is not a
smooth point of the function (the denominator in dy/dx
becomes 0).

. . . . . .

Example

Solution

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

The possible solution x = 0 leads to y = 0, which is not a
smooth point of the function (the denominator in dy/dx
becomes 0).
The possible solution x = − 2 yields y = ± 3√3 .
3
2

. . . . . .

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

. . . . . .

Example

Solution
dx
Tangent lines are vertical when = 0.
dy
Differentiating x implicitly as a function of y gives
dx dx
2y = 3x2 + 2x , so
dy dy

dx 2y
= 2
dy 3x + 2x

. . . . . .

Example

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

dx 2y
= 2
dy 3x + 2x

This is 0 only when y = 0.

. . . . . .

Example

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

dx 2y
= 2
dy 3x + 2x

This is 0 only when y = 0.
We get the false solution x = 0 and the real solution x = −1.

. . . . . .

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 + 2x cos(x2 )

Now factor and divide:
2x(y3 + cos x2 )
y′ =
5y4 + 3x2 y2

. . . . . .

Examples
Example
Show that the families of curves

xy = c x2 − y2 = k

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

. . . . . .

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.
. . . . . .

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
. . . . . .

.

Definition
The isothermic compressibility of a fluid is defined by

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

.
Image credit: Neil Better
. . . . . .

.

Definition
The isothermic compressibility of a fluid is defined by

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

. The smaller the β , the “harder” the fluid.
Image credit: Neil Better
. . . . . .

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.
We have
dy dy p x p −1
yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1
dx dx q y

. . . . . .


Theorem
p p/q−1
q

Proof.
We have
dy dy p x p −1
yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1
dx dx q y

Now yq−1 = x(p/q)(q−1) = xp−p/q so

x p −1
= xp−1−(p−p/q) = xp/q−1
y q −1

. . . . . .

Lesson 11: Implicit Differentiation

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