Lesson 16: Implicit Differentiation
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Lesson 16 (Section 3.6)Implicit Differentiation
Derivatives of Inverse Functions
Math 1a
October 31, 2007
Announcements
I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)
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ProblemFind the slope of the line which is tangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
Solution (Old Way)
y2 = 1− x2
=⇒ y = −√
1− x2
dy
dx= − −2x
2√
1− x2=
x√1− x2
dy
dx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/5
4/5=
3
4.
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ProblemFind the slope of the line which is tangent to the curve
x2 + y2 = 1
at the point (3/5,−4/5).
Solution (Old Way)
y2 = 1− x2
=⇒ y = −√
1− x2
dy
dx= − −2x
2√
1− x2=
x√1− x2
dy
dx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/5
4/5=
3
4.
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Solution (New Way)
Differentiating this, but pretend that y is a function of x. We have:
2x + 2ydy
dx= 0,
ordy
dx= −x
y.
So if x =3
5, y = −4
5, and r = 1, then the slope of the line tangent
to the circle isdy
dx
∣∣∣∣( 3
5,− 4
5 )=
3/5
4/5=
3
4.
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Summary
any time a relation is given between x and y, we may differentiatey as a function of x even though it is not explicitly definedThis is called implicit differentiation.
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Example
Find the equation of the line tangent to the curve
y2 = x2(x + 1) = x3 + x2
at the point (3,−6).
SolutionDifferentiating the expressionimplicitly with respect to x
gives 2ydy
dx= 3x2 + 2x , so
dy
dx=
3x2 + 2x
2y, and
dy
dx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −11
4.
Thus the equation of thetangent line is
y + 6 = −11
4(x − 3).
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Example
Find the equation of the line tangent to the curve
y2 = x2(x + 1) = x3 + x2
at the point (3,−6).
SolutionDifferentiating the expressionimplicitly with respect to x
gives 2ydy
dx= 3x2 + 2x , so
dy
dx=
3x2 + 2x
2y, and
dy
dx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −11
4.
Thus the equation of thetangent line is
y + 6 = −11
4(x − 3).
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Example
Find the equation of the line tangent to the curve
y2 = x2(x + 1) = x3 + x2
at the point (3,−6).
SolutionDifferentiating the expressionimplicitly with respect to x
gives 2ydy
dx= 3x2 + 2x , so
dy
dx=
3x2 + 2x
2y, and
dy
dx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −11
4.
Thus the equation of thetangent line is
y + 6 = −11
4(x − 3).
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Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We need 3x2 + 1 = 0, or x =1√3
. Then
y2 =1
3√
3+
1
3,
Thus
y = ±
√1
3+
1
3√
3
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Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We need 3x2 + 1 = 0, or x =1√3
. Then
y2 =1
3√
3+
1
3,
Thus
y = ±
√1
3+
1
3√
3
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An example from chemistry
Example
The van der Waals equationdescribes nonideal propertiesof a gas:(
P + an2
V 2
)(V−nb) = nRT ,
where P is the pressure, Vthe volume, T thetemperature, n the number ofmoles of the gas, R aconstant, a is a measure ofattraction between particlesof the gas, and b a measureof particle size.
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An example from chemistry
Example
The van der Waals equationdescribes nonideal propertiesof a gas:(
P + an2
V 2
)(V−nb) = nRT ,
where P is the pressure, Vthe volume, T thetemperature, n the number ofmoles of the gas, R aconstant, a is a measure ofattraction between particlesof the gas, and b a measureof particle size.
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DefinitionThe isothermic compressibility of a fluid is defined by
β = −dV
dP
1
V
with temperature held constant.
The smaller the β, the “harder” the fluid.
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Math 1a - October 31, 2007.GWBWednesday, Oct 31, 2007
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DefinitionThe isothermic compressibility of a fluid is defined by
β = −dV
dP
1
V
with temperature held constant.
The smaller the β, the “harder” the fluid.
![Page 16: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/16.jpg)
Let’s find the compressibility of a van der Waals gas.Differentiating the van der Waals equation by treating V as afunction of P gives(
P +an2
V 2
)dV
dP+ (V − bn)
(1− 2an2
V 3
dV
dP
)= 0,
so
β = − 1
V
dV
dP=
V 2(V − nb)
2abn3 − an2V + PV 3
I What if a = b = 0?
I What is the sign ofdβ
db?
I What is the sign ofdβ
da?
![Page 17: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/17.jpg)
Let’s find the compressibility of a van der Waals gas.Differentiating the van der Waals equation by treating V as afunction of P gives(
P +an2
V 2
)dV
dP+ (V − bn)
(1− 2an2
V 3
dV
dP
)= 0,
so
β = − 1
V
dV
dP=
V 2(V − nb)
2abn3 − an2V + PV 3
I What if a = b = 0?
I What is the sign ofdβ
db?
I What is the sign ofdβ
da?
![Page 18: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/18.jpg)
Let’s find the compressibility of a van der Waals gas.Differentiating the van der Waals equation by treating V as afunction of P gives(
P +an2
V 2
)dV
dP+ (V − bn)
(1− 2an2
V 3
dV
dP
)= 0,
so
β = − 1
V
dV
dP=
V 2(V − nb)
2abn3 − an2V + PV 3
I What if a = b = 0?
I What is the sign ofdβ
db?
I What is the sign ofdβ
da?
![Page 19: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/19.jpg)
Let’s find the compressibility of a van der Waals gas.Differentiating the van der Waals equation by treating V as afunction of P gives(
P +an2
V 2
)dV
dP+ (V − bn)
(1− 2an2
V 3
dV
dP
)= 0,
so
β = − 1
V
dV
dP=
V 2(V − nb)
2abn3 − an2V + PV 3
I What if a = b = 0?
I What is the sign ofdβ
db?
I What is the sign ofdβ
da?
![Page 20: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/20.jpg)
Let’s find the compressibility of a van der Waals gas.Differentiating the van der Waals equation by treating V as afunction of P gives(
P +an2
V 2
)dV
dP+ (V − bn)
(1− 2an2
V 3
dV
dP
)= 0,
so
β = − 1
V
dV
dP=
V 2(V − nb)
2abn3 − an2V + PV 3
I What if a = b = 0?
I What is the sign ofdβ
db?
I What is the sign ofdβ
da?
![Page 21: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/21.jpg)
Math 1a - October 31, 2007.GWBWednesday, Oct 31, 2007
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Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f ′(a) 6= 0. Then f −1 is defined ina neighborhood of b = f (a), and
(f −1)′(b) =1
f ′(f −1(b))
“Proof”.If y = f (x), take f −1 of both sides to get
x = f −1(f (x)),
so1 = (f −1)′(f (x))f ′(x).
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Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f ′(a) 6= 0. Then f −1 is defined ina neighborhood of b = f (a), and
(f −1)′(b) =1
f ′(f −1(b))
“Proof”.If y = f (x), take f −1 of both sides to get
x = f −1(f (x)),
so1 = (f −1)′(f (x))f ′(x).
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Using implicit differentiation to find derivatives
Example
Finddy
dxif y =
√x .
SolutionIf y =
√x, then
y2 = x ,
so
2ydy
dx= 1 =⇒ dy
dx=
1
2y=
1
2√
x.
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Using implicit differentiation to find derivatives
Example
Finddy
dxif y =
√x .
SolutionIf y =
√x, then
y2 = x ,
so
2ydy
dx= 1 =⇒ dy
dx=
1
2y=
1
2√
x.
![Page 26: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/26.jpg)
The power rule for rational numbers
Example
Finddy
dxif y = xp/q, where p and q are integers.
SolutionWe have
yq = xp
qyq−1 dy
dx= pxp−1
dy
dx=
p
q· 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
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The power rule for rational numbers
Example
Finddy
dxif y = xp/q, where p and q are integers.
SolutionWe have
yq = xp
qyq−1 dy
dx= pxp−1
dy
dx=
p
q· 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
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The power rule for rational numbers
Example
Finddy
dxif y = xp/q, where p and q are integers.
SolutionWe have
yq = xp
qyq−1 dy
dx= pxp−1
dy
dx=
p
q· 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
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Arcsin
Remember that
arcsin x = sin−1 x 6= 1
sin x
Example
Find the derivative of arcsin.
SolutionWe have y = arcsin x =⇒ x = sin y , so
1 = cos ydy
dx=⇒ dy
dx=
1
cos y.
We would like to express this in terms of x. Since x = sin y, wehave
cos y =√
1− x2 =⇒ d
dxarcsin x =
1√1− x2
.
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Arcsin
Remember that
arcsin x = sin−1 x 6= 1
sin x
Example
Find the derivative of arcsin.
SolutionWe have y = arcsin x =⇒ x = sin y , so
1 = cos ydy
dx=⇒ dy
dx=
1
cos y.
We would like to express this in terms of x. Since x = sin y, wehave
cos y =√
1− x2 =⇒ d
dxarcsin x =
1√1− x2
.
![Page 31: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/31.jpg)
Math 1a - October 31, 2007.GWBWednesday, Oct 31, 2007
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Arcsin
Remember that
arcsin x = sin−1 x 6= 1
sin x
Example
Find the derivative of arcsin.
SolutionWe have y = arcsin x =⇒ x = sin y , so
1 = cos ydy
dx=⇒ dy
dx=
1
cos y.
We would like to express this in terms of x. Since x = sin y, wehave
cos y =√
1− x2 =⇒ d
dxarcsin x =
1√1− x2
.
![Page 33: Lesson 16: Implicit Differentiation](https://reader034.fdocuments.us/reader034/viewer/2022052523/556c3cf2d8b42a02258b5743/html5/thumbnails/33.jpg)
Arccos
Example
Find the derivative of arccos.
Answer
d
dx(arccos x) =
−1√1− x2
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Arccos
Example
Find the derivative of arccos.
Answer
d
dx(arccos x) =
−1√1− x2
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Math 1a - October 31, 2007.GWBWednesday, Oct 31, 2007
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Arctan
Example
Find the derivative of arctan.
SolutionFrom y = arctan x we have
x = tan y =⇒ 1 = sec2 ydy
dx,
sody
dx=
1
sec2 y.
But for all y
1 + tan2 y = sec2 y , =⇒ dy
dx=
1
1 + x2.
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Arctan
Example
Find the derivative of arctan.
SolutionFrom y = arctan x we have
x = tan y =⇒ 1 = sec2 ydy
dx,
sody
dx=
1
sec2 y.
But for all y
1 + tan2 y = sec2 y , =⇒ dy
dx=
1
1 + x2.
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Arctan
Example
Find the derivative of arctan.
SolutionFrom y = arctan x we have
x = tan y =⇒ 1 = sec2 ydy
dx,
sody
dx=
1
sec2 y.
But for all y
1 + tan2 y = sec2 y , =⇒ dy
dx=
1
1 + x2.
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Arcsec
Example
Find the derivative of arcsec.
SolutionIf y = arcsec x then x = sec y, so
1 = sec y tan ydy
dx.
Thusdy
dx=
1
x tan y=
1
x√
x2 − 1.
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Arcsec
Example
Find the derivative of arcsec.
SolutionIf y = arcsec x then x = sec y, so
1 = sec y tan ydy
dx.
Thusdy
dx=
1
x tan y=
1
x√
x2 − 1.