The document is a lesson on implicit differentiation and related concepts: 1) Implicit differentiation allows one to take the derivative of an implicitly defined relation between x and y, even if y is not explicitly defined as a function of x. 2) Examples are provided to demonstrate implicit differentiation, such as finding the slope of a tangent line to a curve. 3) The van der Waals equation is introduced to describe non-ideal gas properties, and implicit differentiation is used to find the isothermal compressibility of a van der Waals gas.

1. Lesson 16 (Section 3.6)
Implicit Differentiation
Derivatives of Inverse Functions
Math 1a
October 31, 2007
Announcements
OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)

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

3. Problem
Find 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 − x 2
−2x
dy x
=− √ =√
dx 2 1 − x2
2 1−x
dy 3/5 3/5 3
= = =.
dx x=3/5 4/5 4
1 − (3/5)2

4. Solution (New Way)
Differentiating this, but pretend that y is a function of x. We have:
dy
2x + 2y = 0,
dx
or
dy x
=− .
dx y
3 4
So if x = , y = − , and r = 1, then the slope of the line tangent
5 5
to the circle is
dy 3/5 3
= =.
dx ( 3 ,− 4 ) 4/5 4
5 5

5. Summary
any time a relation is given between x and y , we may differentiate
y as a function of x even though it is not explicitly defined
This is called implicit differentiation.

6. 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).

7. 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
dy
= 3x 2 + 2x, so
gives 2y
dx
3x 2 + 2x
dy
= , and
dx 2y
3 · 32 + 2 · 3
dy 11
=− .
=
dx 2(−6) 4
(3,−6)

8. 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
dy
= 3x 2 + 2x, so
gives 2y
dx
3x 2 + 2x
dy
= , and
dx 2y
3 · 32 + 2 · 3
dy 11
=− .
=
Thus the equation of the
dx 2(−6) 4
(3,−6)
tangent line is
11
y +6=− (x − 3).
4

9. Example
Find the horizontal tangent lines to the same curve: y 2 = x 3 + x 2

10. Example
Find the horizontal tangent lines to the same curve: y 2 = x 3 + x 2
Solution
1
We need 3x 2 + 1 = 0, or x = √ . Then
3
1 1
y2 = √ + ,
33 3
Thus
1 1
+√
y =±
3 33

11. An example from chemistry
Example
The van der Waals equation
describes nonideal properties
of a gas:
n2
(V −nb) = nRT ,
P +a
V2
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.

12. An example from chemistry
Example
The van der Waals equation
describes nonideal properties
of a gas:
n2
(V −nb) = nRT ,
P +a
V2
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.

13. Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.

15. 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.

16. 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 2an2 dV
dV
+ (V − bn) 1 −
P+ = 0,
V2 V 3 dP
dP

17. 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 2an2 dV
dV
+ (V − bn) 1 −
P+ = 0,
V2 V 3 dP
dP
so
V 2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV 3
V dP

18. 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 2an2 dV
dV
+ (V − bn) 1 −
P+ = 0,
V2 V 3 dP
dP
so
V 2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV 3
V dP
What if a = b = 0?

19. 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 2an2 dV
dV
+ (V − bn) 1 −
P+ = 0,
V2 V 3 dP
dP
so
V 2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV 3
V dP
What if a = b = 0?
dβ
What is the sign of ?
db

20. 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 2an2 dV
dV
+ (V − bn) 1 −
P+ = 0,
V2 V 3 dP
dP
so
V 2 (V − nb)
1 dV
β=− =
2abn3 − an2 V + PV 3
V dP
What if a = b = 0?
dβ
What is the sign of ?
db
dβ
What is the sign of ?
da

22. Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f (a) = 0. Then f −1 is defined in
a neighborhood of b = f (a), and
1
(f −1 ) (b) =
(f −1 (b))
f

23. Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f (a) = 0. Then f −1 is defined in
a neighborhood of b = f (a), and
1
(f −1 ) (b) =
(f −1 (b))
f
“Proof”.
If y = f (x), take f −1 of both sides to get
x = f −1 (f (x)),
so
1 = (f −1 ) (f (x))f (x).

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

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

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

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

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

29. Arcsin
Remember that
1
arcsin x = sin−1 x =
sin x

30. Arcsin
Remember that
1
arcsin x = sin−1 x =
sin x
Example
Find the derivative of arcsin.

32. Arcsin
Remember that
1
arcsin x = sin−1 x =
sin x
Example
Find the derivative of arcsin.
Solution
We have y = arcsin x =⇒ x = sin y , so
dy dy 1
1 = cos y =⇒ = .
dx dx cos y
We would like to express this in terms of x. Since x = sin y , we
have
d 1
cos y = 1 − x 2 =⇒ arcsin x = √ .
dx 1 − x2

33. Arccos
Example
Find the derivative of arccos.

34. Arccos
Example
Find the derivative of arccos.
Answer
−1
d
(arccos x) = √
dx 1 − x2

36. Arctan
Example
Find the derivative of arctan.

37. Arctan
Example
Find the derivative of arctan.
Solution
From y = arctan x we have
dy
x = tan y =⇒ 1 = sec2 y ,
dx
so
dy 1
= .
sec2 y
dx

38. Arctan
Example
Find the derivative of arctan.
Solution
From y = arctan x we have
dy
x = tan y =⇒ 1 = sec2 y ,
dx
so
dy 1
= .
sec2 y
dx
But for all y
dy 1
1 + tan2 y = sec2 y , =⇒ = .
1 + x2
dx

39. Arcsec
Example
Find the derivative of arcsec.

40. Arcsec
Example
Find the derivative of arcsec.
Solution
If y = arcsec x then x = sec y , so
dy
1 = sec y tan y .
dx
Thus
dy 1 1
=√
= .
dx x tan y x x2 − 1