Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.

1. 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
. . . . . .

2. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .

3. 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).
. . . . . .

4. 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).
. . . . . .

5. 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). .
. . . . . .

6. 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). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
. . . . . .

7. 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). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2
. . . . . .

8. 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). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
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
. . . . . .

9. 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). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
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
. . . . . .

10. 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)
. . . . . .

11. 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.
. . . . . .

12. 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.
. . . . . .

13. 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.
l
.ooks like a function
. . . . . .

14. 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.
. . . . . .

15. 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.
. . . . . .

16. 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.
l
.ooks like a function
So f(x) is defined
“locally” and is . x
.
differentiable
The chain rule then
applies for this local
choice.
. . . . . .

17. 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.
. . . . . .

18. 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.
. . . . . .

19. 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 does not look like a
applies for this local function, but that’s
choice. OK—there are only
two points like this
. . . . . .

20. 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
. . . . . .

21. 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.
. . . . . .

22. Mnemonic
Explicit Implicit
y = f(x) F(x, y) = k
. . . . . .

23. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .

24. 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).
.
. . . . . .

25. 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).
.
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
. . . . . .

26. 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).
.
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
. . . . . .

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

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

29. Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We solve for dy/dx = 0:
3x2 + 2x
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
2y
. . . . . .

30. Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We solve for dy/dx = 0:
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).
. . . . . .

31. Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We solve for dy/dx = 0:
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
. . . . . .

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

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

34. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
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
This is 0 only when y = 0.
. . . . . .

35. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
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
This is 0 only when y = 0.
We get the false solution x = 0 and the real solution x = −1.
. . . . . .

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

37. 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
. . . . . .

38. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
. . . . . .

39. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
In the first curve,
y
y + xy′ = 0 =⇒ y′ = −
x
. . . . . .

40. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
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.
. . . . . .

41. 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
. . . . . .

42. .
Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.
.
Image credit: Neil Better
. . . . . .

43. .
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
. . . . . .

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

45. 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.
. . . . . .

46. 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.
. . . . . .

47. 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
. . . . . .

48. 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
. . . . . .

49. 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
so
1 dV V2 (V − nb)
β=− =
V dP 2abn3 − an2 V + PV3
. . . . . .

50. 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
so
1 dV V2 (V − nb)
β=− =
V dP 2abn3 − an2 V + PV3
What if a = b = 0?
. . . . . .

51. 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
so
1 dV V2 (V − nb)
β=− =
V dP 2abn3 − an2 V + PV3
What if a = b = 0?
dβ
Without taking the derivative, what is the sign of ?
db
. . . . . .

52. 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
so
1 dV V2 (V − nb)
β=− =
V dP 2abn3 − an2 V + PV3
What if a = b = 0?
dβ
Without taking the derivative, what is the sign of ?
db
dβ
Without taking the derivative, what is the sign of ?
da
. . . . . .

53. 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).
. . . . . .

54. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .

55. Using implicit differentiation to find derivatives
Example
dy √
Find if y = x.
dx
. . . . . .

56. 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
. . . . . .

57. 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
. . . . . .

58. 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
Proof.
We have
dy dy p x p −1
yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1
dx dx q y
. . . . . .

59. 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
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
. . . . . .