This document discusses implicit differentiation and finding the slope of tangent lines using implicit differentiation. It begins with an example problem of finding the slope of the tangent line to the curve x^2 + y^2 = 1 at the point (3/5, -4/5). It then explains how to set up and solve the implicit differentiation problem to find the slope. The document emphasizes that even when a relation is not explicitly a function, it can often be treated as locally functional to apply implicit differentiation and find tangent slopes. It provides another example problem and discusses horizontal and vertical tangent lines.

1. Section 2.6
Implicit Differentiation
V63.0121, Calculus I
February 24/25, 2009
Announcements
Midterm in class March 4/5
ALEKS due Friday, 11:59pm
.
.
Image credit: Telstar Logistics
. . . . . .

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

3. y
.
Motivating Example
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. y
.
Motivating Example
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. y
.
Motivating Example
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. y
.
Motivating Example
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. y
.
Motivating Example
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 −?)
−2x
dy x
=− √ =√
Differentiate:
2 1 − x2 1 − x2
dx
. . . . . .

8. y
.
Motivating Example
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 −?)
−2x
dy x
=− √ =√
Differentiate:
2 1 − x2 1 − x2
dx
dy 3/5 3/5 3
=√ = =.
Evaluate:
1 − (3/5)
dx x=3/5 4/5 4
2
. . . . . .

9. y
.
Motivating Example
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 −?)
−2x
dy x
=− √ =√
Differentiate:
2 1 − x2 1 − x2
dx
dy 3/5 3/5 3
=√ = =.
Evaluate:
1 − (3/5)
dx x=3/5 4/5 4
2
. . . . . .

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!
y
.
“Near” most points on
the curve x2 + y2 = 1, the
curve resembles the
graph of a function.
So f(x) is defined “locally”
. x
.
and is differentiable
The chain rule then
applies for this local
.
choice.
. . . . . .

12. 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.
So f(x) is defined “locally”
. x
.
and is differentiable
The chain rule then
applies for this local
.
choice.
. . . . . .

13. 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.
So f(x) is defined “locally”
. x
.
and is 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!
y
.
“Near” most points on
the curve x2 + y2 = 1, the .
curve resembles the
graph of a function.
So f(x) is defined “locally”
. x
.
and is differentiable
The chain rule then
applies for this local
choice.
. . . . . .

15. 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.
So f(x) is defined “locally”
. x
.
and is differentiable
The chain rule then
applies for this local
choice.
. . . . . .

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

17. 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.
So f(x) is defined “locally”
. . x
.
and is differentiable
The chain rule then
applies for this local
choice.
. . . . . .

18. 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.
So f(x) is defined “locally”
. . x
.
and is differentiable
The chain rule then
applies for this local
choice.
. . . . . .

19. 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.
So f(x) is defined “locally”
. . x
.
and is 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
y
.
If a relation is given between x and
y,
“Most of the time” “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
.
.
Image credit: Walsh
. . . . . .

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

24. Example
Find the equation of the line
tangent to the curve
.
y2 = x2 (x + 1) = x3 + x2
at the point (3, −6).
.
. . . . . .

25. Example
Find the equation of the line
tangent to the curve
.
y2 = x2 (x + 1) = x3 + x2
at the point (3, −6).
.
Solution
Differentiating the expression implicitly with respect to x gives
3x2 + 2x
dy dy
2y = 3x2 + 2x, so = , and
dx dx 2y
3 · 32 + 2 · 3
dy 11
=− .
=
dx 2(−6) 4
(3,−6)
. . . . . .

26. Example
Find the equation of the line
tangent to the curve
.
y2 = x2 (x + 1) = x3 + x2
at the point (3, −6).
.
Solution
Differentiating the expression implicitly with respect to x gives
3x2 + 2x
dy dy
2y = 3x2 + 2x, so = , and
dx dx 2y
3 · 32 + 2 · 3
dy 11
=− .
=
dx 2(−6) 4
(3,−6)
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 .
2
3
. . . . . .

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

34. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
dx
= 0.
Tangent lines are vertical when
dy
Differentiating x implicitly as a function of y gives
dx dx
2y = 3x2 + 2x , so
dy dy
dx 2y
=2
3x + 2x
dy
This is 0 only when y = 0.
. . . . . .

35. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
dx
= 0.
Tangent lines are vertical when
dy
Differentiating x implicitly as a function of y gives
dx dx
2y = 3x2 + 2x , so
dy dy
dx 2y
=2
3x + 2x
dy
This is 0 only when y = 0.
We get the false solution x = 0 and the real solution x = −1.
. . . . . .

36. 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
. . . . . .

37. .
Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.
.
Image credit: Neil Better
. . . . . .

38. .
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
. . . . . .

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

40. 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.
. . . . . .

41. 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
..
O. H
. ydrogen bonds
. xygen
where P is the pressure, V the
H
..
volume, T the temperature, n .
the number of moles of the
O.
. xygen . .
H
gas, R a constant, a is a
measure of attraction between
H
..
particles of the gas, and b a
measure of particle size.
. . . . . .

42. 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
. . . . . .

43. 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
+ (V − bn) 1 − 3
P+ 2 = 0,
dP V dP
V
. . . . . .

44. 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
+ (V − bn) 1 − 3
P+ 2 = 0,
dP V dP
V
so
V2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV3
V dP
. . . . . .

45. 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
+ (V − bn) 1 − 3
P+ 2 = 0,
dP V dP
V
so
V2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV3
V dP
What if a = b = 0?
. . . . . .

46. 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
+ (V − bn) 1 − 3
P+ 2 = 0,
dP V dP
V
so
V2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV3
V dP
What if a = b = 0?
dβ
Without taking the derivative, what is the sign of ?
db
. . . . . .

47. 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
+ (V − bn) 1 − 3
P+ 2 = 0,
dP V dP
V
so
V2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV3
V dP
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
. . . . . .

48. Nasty derivatives
(2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 )
dβ
=−
(2abn3 − an2 V + PV3 )2
db
(2 )
nV3 an + PV2
= −( )2 < 0
PV3 + an2 (2bn − V)
n2 (bn − V)(2bn − V)V2
dβ
=( )2 > 0
da PV3 + an2 (2bn − V)
(as long as V > 2nb, and it’s probably true that V ≫ 2nb).
. . . . . .

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

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

51. Using implicit differentiation to find derivatives
Example
√
dy
if y = x.
Find
dx
Solution
√
If y = x, then
y2 = x,
so
dy dy 1 1
= √.
= 1 =⇒ =
2y
dx dx 2y 2x
. . . . . .

52. The power rule for rational numbers
Example
dy
if y = xp/q , where p and q are integers.
Find
dx
. . . . . .

53. The power rule for rational numbers
Example
dy
if y = xp/q , where p and q are integers.
Find
dx
Solution
We have
p xp−1
dy dy
= · q−1
yq = xp =⇒ qyq−1 = pxp−1 =⇒
dx dx qy
. . . . . .

54. The power rule for rational numbers
Example
dy
if y = xp/q , where p and q are integers.
Find
dx
Solution
We have
p xp−1
dy dy
= · q−1
yq = xp =⇒ qyq−1 = pxp−1 =⇒
dx dx qy
Now yq−1 = xp(q−1)/q = xp−p/q so
xp−1
= xp−1−(p−p/q) = xp/q−1
yq−1
. . . . . .