lesson 12: implicit differentiation
DESCRIPTION
A functional relationship between y and x can be made explicit. Even if the relation is not functional, however, we can assume the relation usually defines y as a function.TRANSCRIPT
. . . . . .
Section 2.6Implicit Differentiation
V63.0121, Calculus I
February 24/25, 2009
Announcements
I Midterm in class March 4/5I ALEKS due Friday, 11:59pm
..Image credit: Telstar Logistics
. . . . . .
Outline
The big idea, by example
ExamplesVertical and Horizontal TangentsChemistry
The power rule for rational powers
. . . . . .
Motivating Example
ProblemFind the slope of the line which istangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.Solution (Explicit)
I Isolate: y2 = 1 − x2 =⇒ y = −√
1 − x2. (Why the −?)
I Differentiate:dydx
= − −2x
2√
1 − x2=
x√1 − x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1 − (3/5)2=
3/54/5
=34.
. . . . . .
Motivating Example
ProblemFind the slope of the line which istangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.Solution (Explicit)
I Isolate: y2 = 1 − x2 =⇒ y = −√
1 − x2. (Why the −?)
I Differentiate:dydx
= − −2x
2√
1 − x2=
x√1 − x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1 − (3/5)2=
3/54/5
=34.
. . . . . .
Motivating Example
ProblemFind the slope of the line which istangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1 − x2 =⇒ y = −√
1 − x2. (Why the −?)
I Differentiate:dydx
= − −2x
2√
1 − x2=
x√1 − x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1 − (3/5)2=
3/54/5
=34.
. . . . . .
Motivating Example
ProblemFind the slope of the line which istangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.Solution (Explicit)
I Isolate: y2 = 1 − x2 =⇒ y = −√
1 − x2. (Why the −?)
I Differentiate:dydx
= − −2x
2√
1 − x2=
x√1 − x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1 − (3/5)2=
3/54/5
=34.
. . . . . .
Motivating Example
ProblemFind the slope of the line which istangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.Solution (Explicit)
I Isolate: y2 = 1 − x2 =⇒ y = −√
1 − x2. (Why the −?)
I Differentiate:dydx
= − −2x
2√
1 − x2=
x√1 − x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1 − (3/5)2=
3/54/5
=34.
. . . . . .
Motivating Example
ProblemFind the slope of the line which istangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.Solution (Explicit)
I Isolate: y2 = 1 − x2 =⇒ y = −√
1 − x2. (Why the −?)
I Differentiate:dydx
= − −2x
2√
1 − x2=
x√1 − x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1 − (3/5)2=
3/54/5
=34.
. . . . . .
Motivating Example
ProblemFind the slope of the line which istangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.Solution (Explicit)
I Isolate: y2 = 1 − x2 =⇒ y = −√
1 − x2. (Why the −?)
I Differentiate:dydx
= − −2x
2√
1 − x2=
x√1 − x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1 − (3/5)2=
3/54/5
=34.
. . . . . .
We know that x2 + y2 = 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) = − xf(x)
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
.looks like a function
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
.
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
.
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
.
.looks like a function
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
.
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
.
. . . . . .
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points onthe curve x2 + y2 = 1, thecurve resembles thegraph of a function.
I So f(x) is defined “locally”and is differentiable
I The chain rule thenapplies for this localchoice.
. .x
.y
.
.
.does not look like afunction, but that’s
OK—there are onlytwo points like this
.
. . . . . .
ProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 at thepoint (3/5,−4/5).
Solution (Implicit, with Leibniz notation)
I Differentiate. Remember y is assumed to be a function of x:
2x + 2ydydx
= 0,
I Isolatedydx
:
dydx
= −xy.
I Evaluate:dydx
∣∣∣∣( 3
5 ,− 45)
=3/54/5
=34.
. . . . . .
Summary
If a relation is given between x andy,
I “Most of the time” “at mostplaces” y can be assumed tobe a function of x
I we may differentiate therelation as is
I Solving fordydx
does give the
slope of the tangent line tothe curve at a point on thecurve.
. .x
.y
.
. . . . . .
Mnemonic
Explicit Implicit
y = f(x) F(x, y) = k
.
.Image credit: Walsh
. . . . . .
Outline
The big idea, by example
ExamplesVertical and Horizontal TangentsChemistry
The power rule for rational powers
. . . . . .
ExampleFind the equation of the linetangent to the curve
y2 = x2(x + 1) = x3 + x2
at the point (3,−6).
.
.
SolutionDifferentiating the expression implicitly with respect to x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −11
4.
Thus the equation of the tangent line is y + 6 = −114
(x − 3).
. . . . . .
ExampleFind the equation of the linetangent to the curve
y2 = x2(x + 1) = x3 + x2
at the point (3,−6).
.
.SolutionDifferentiating the expression implicitly with respect to x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −11
4.
Thus the equation of the tangent line is y + 6 = −114
(x − 3).
. . . . . .
ExampleFind the equation of the linetangent to the curve
y2 = x2(x + 1) = x3 + x2
at the point (3,−6).
.
.SolutionDifferentiating the expression implicitly with respect to x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −11
4.
Thus the equation of the tangent line is y + 6 = −114
(x − 3).
. . . . . .
ExampleFind the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
I We solve for dy/dx = 0:
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
I The possible solution x = 0 leads to y = 0, which is not a smoothpoint of the function (the denominator in dy/dx becomes 0).
I The possible solution x = − 23 yields y = ± 2
3√
3.
. . . . . .
ExampleFind the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
I We solve for dy/dx = 0:
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
I The possible solution x = 0 leads to y = 0, which is not a smoothpoint of the function (the denominator in dy/dx becomes 0).
I The possible solution x = − 23 yields y = ± 2
3√
3.
. . . . . .
ExampleFind the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
I We solve for dy/dx = 0:
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
I The possible solution x = 0 leads to y = 0, which is not a smoothpoint of the function (the denominator in dy/dx becomes 0).
I The possible solution x = − 23 yields y = ± 2
3√
3.
. . . . . .
ExampleFind the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
I We solve for dy/dx = 0:
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
I The possible solution x = 0 leads to y = 0, which is not a smoothpoint of the function (the denominator in dy/dx becomes 0).
I The possible solution x = − 23 yields y = ± 2
3√
3.
. . . . . .
ExampleFind the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
I We solve for dy/dx = 0:
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
I The possible solution x = 0 leads to y = 0, which is not a smoothpoint of the function (the denominator in dy/dx becomes 0).
I The possible solution x = − 23 yields y = ± 2
3√
3.
. . . . . .
ExampleFind the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2 dxdy
+ 2xdxdy
, so
dxdy
=2y
3x2 + 2x
I This is 0 only when y = 0.I We get the false solution x = 0 and the real solution x = −1.
. . . . . .
ExampleFind the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2 dxdy
+ 2xdxdy
, so
dxdy
=2y
3x2 + 2x
I This is 0 only when y = 0.I We get the false solution x = 0 and the real solution x = −1.
. . . . . .
ExampleFind the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2 dxdy
+ 2xdxdy
, so
dxdy
=2y
3x2 + 2x
I This is 0 only when y = 0.
I We get the false solution x = 0 and the real solution x = −1.
. . . . . .
ExampleFind the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2 dxdy
+ 2xdxdy
, so
dxdy
=2y
3x2 + 2x
I This is 0 only when y = 0.I We get the false solution x = 0 and the real solution x = −1.
. . . . . .
Ideal gases
The ideal gas law relatestemperature, pressure, andvolume of a gas:
PV = nRT
(R is a constant, n is theamount of gas in moles)
.
.Image credit: Scott Beale / Laughing Squid
. . . . . .
.
.Image credit: Neil Better
DefinitionThe isothermic compressibility of a fluid is defined by
β = −dVdP
1V
with temperature held constant.
The smaller the β, the “harder” the fluid.
. . . . . .
.
.Image credit: Neil Better
DefinitionThe isothermic compressibility of a fluid is defined by
β = −dVdP
1V
with temperature held constant.
The smaller the β, the “harder” the fluid.
. . . . . .
ExampleFind the isothermic compressibility of an ideal gas.
SolutionIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then
dPdP
· V + PdVdP
= 0 =⇒ dVdP
= −VP
So
β = −1V· dV
dP=
1P
Compressibility and pressure are inversely related.
. . . . . .
ExampleFind the isothermic compressibility of an ideal gas.
SolutionIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then
dPdP
· V + PdVdP
= 0 =⇒ dVdP
= −VP
So
β = −1V· dV
dP=
1P
Compressibility and pressure are inversely related.
. . . . . .
Nonideal gassesNot that there’s anything wrong with that
ExampleThe van der Waals equationmakes fewer simplifications:(
P + an2
V2
)(V − nb) = nRT,
where P is the pressure, V thevolume, T the temperature, nthe number of moles of thegas, R a constant, a is ameasure of attraction betweenparticles of the gas, and b ameasure of particle size.
...Oxygen
..H
..H
..Oxygen
..H
..H
..Oxygen ..H
..H
.
.
.Hydrogen bonds
. . . . . .
Nonideal gassesNot that there’s anything wrong with that
ExampleThe van der Waals equationmakes fewer simplifications:(
P + an2
V2
)(V − nb) = nRT,
where P is the pressure, V thevolume, T the temperature, nthe number of moles of thegas, R a constant, a is ameasure of attraction betweenparticles of the gas, and b ameasure of particle size. .
.Image credit: Wikimedia Commons
. . . . . .
Let’s find the compressibility of a van der Waals gas. Differentiatingthe van der Waals equation by treating V as a function of P gives(
P +an2
V2
)dVdP
+ (V − bn)(
1 − 2an2
V3
dVdP
)= 0,
so
β = −1V
dVdP
=V2(V − nb)
2abn3 − an2V + PV3
I What if a = b = 0?
I Without taking the derivative, what is the sign ofdβdb
?
I Without taking the derivative, what is the sign ofdβda
?
. . . . . .
Let’s find the compressibility of a van der Waals gas. Differentiatingthe van der Waals equation by treating V as a function of P gives(
P +an2
V2
)dVdP
+ (V − bn)(
1 − 2an2
V3
dVdP
)= 0,
so
β = −1V
dVdP
=V2(V − nb)
2abn3 − an2V + PV3
I What if a = b = 0?
I Without taking the derivative, what is the sign ofdβdb
?
I Without taking the derivative, what is the sign ofdβda
?
. . . . . .
Let’s find the compressibility of a van der Waals gas. Differentiatingthe van der Waals equation by treating V as a function of P gives(
P +an2
V2
)dVdP
+ (V − bn)(
1 − 2an2
V3
dVdP
)= 0,
so
β = −1V
dVdP
=V2(V − nb)
2abn3 − an2V + PV3
I What if a = b = 0?
I Without taking the derivative, what is the sign ofdβdb
?
I Without taking the derivative, what is the sign ofdβda
?
. . . . . .
Let’s find the compressibility of a van der Waals gas. Differentiatingthe van der Waals equation by treating V as a function of P gives(
P +an2
V2
)dVdP
+ (V − bn)(
1 − 2an2
V3
dVdP
)= 0,
so
β = −1V
dVdP
=V2(V − nb)
2abn3 − an2V + PV3
I What if a = b = 0?
I Without taking the derivative, what is the sign ofdβdb
?
I Without taking the derivative, what is the sign ofdβda
?
. . . . . .
Let’s find the compressibility of a van der Waals gas. Differentiatingthe van der Waals equation by treating V as a function of P gives(
P +an2
V2
)dVdP
+ (V − bn)(
1 − 2an2
V3
dVdP
)= 0,
so
β = −1V
dVdP
=V2(V − nb)
2abn3 − an2V + PV3
I What if a = b = 0?
I Without taking the derivative, what is the sign ofdβdb
?
I Without taking the derivative, what is the sign ofdβda
?
. . . . . .
Nasty derivatives
I
dβdb
= −(2abn3 − an2V + PV3)(nV2) − (nbV2 − V3)(2an3)
(2abn3 − an2V + PV3)2
= −nV3 (
an2 + PV2)(PV3 + an2(2bn − V)
)2 < 0
Idβda
=n2(bn − V)(2bn − V)V2(PV3 + an2(2bn − V)
)2 > 0
(as long as V > 2nb, and it’s probably true that V ≫ 2nb).
. . . . . .
Outline
The big idea, by example
ExamplesVertical and Horizontal TangentsChemistry
The power rule for rational powers
. . . . . .
Using implicit differentiation to find derivatives
Example
Finddydx
if y =√
x.
SolutionIf y =
√x, then
y2 = x,
so
2ydydx
= 1 =⇒ dydx
=12y
=1
2√
x.
. . . . . .
Using implicit differentiation to find derivatives
Example
Finddydx
if y =√
x.
SolutionIf y =
√x, then
y2 = x,
so
2ydydx
= 1 =⇒ dydx
=12y
=1
2√
x.
. . . . . .
The power rule for rational numbers
Example
Finddydx
if y = xp/q, where p and q are integers.
SolutionWe have
yq = xp =⇒ qyq−1 dydx
= pxp−1 =⇒ dydx
=pq· xp−1
yq−1
Now yq−1 = xp(q−1)/q = xp−p/q so
xp−1
yq−1 = xp−1−(p−p/q) = xp/q−1
. . . . . .
The power rule for rational numbers
Example
Finddydx
if y = xp/q, where p and q are integers.
SolutionWe have
yq = xp =⇒ qyq−1 dydx
= pxp−1 =⇒ dydx
=pq· xp−1
yq−1
Now yq−1 = xp(q−1)/q = xp−p/q so
xp−1
yq−1 = xp−1−(p−p/q) = xp/q−1
. . . . . .
The power rule for rational numbers
Example
Finddydx
if y = xp/q, where p and q are integers.
SolutionWe have
yq = xp =⇒ qyq−1 dydx
= pxp−1 =⇒ dydx
=pq· xp−1
yq−1
Now yq−1 = xp(q−1)/q = xp−p/q so
xp−1
yq−1 = xp−1−(p−p/q) = xp/q−1