Mixed Partial Derivatives
2
2
∂ f
∂ f
and ∂x∂y
are equal at points
In these notes we prove that the mixed partial derivatives ∂y∂x
where both are continuous. This result goes under several different names including “equality
of mixed partials” and “Clairaut’s theorem”. Following the proof there is an example which
∂2 f
∂2 f
shows that, when ∂y∂x
and ∂x∂y
are not continuous, they can be different.
Theorem 1.
If the partial derivatives
∂2 f
∂y∂x
and
∂2 f
∂x∂y
exist and are continuous at (a, b), then
∂2 f
∂2 f
(a, b) =
(a, b)
∂y∂x
∂x∂y
Proof. Here is an outline of the proof. The details are given as “footnotes” at the end of the
outline. Fix a and b and define(1)
F (h, k) =
1 f (a + h, b + k) − f (a, b + k) − f (a + h, b) + f (a, b)
hk
Then, by the mean value theorem,
1 h ∂f
(a + h, b + θ1 k) −
h ∂y
3 ∂ ∂f
(a + θ2 h, b + θ1 k)
=
∂x ∂y
h
4 1 ∂f
F (h, k) =
(a + θ3 h, b + k) −
k ∂x
5 ∂ ∂f
(a + θ3 h, b + θ4 k)
=
∂y ∂x
2
F (h, k) =
i
∂f
(a, b + θ1 k)
∂y
i
∂f
(a + θ3 h, b)
∂x
2
for some 0 < θ1 , θ2 , θ3 , θ4 < 1. Here “=” means that the equality is shown in “footnote”
number 2. All of θ1 , θ2 , θ3 , θ4 depend on a, b, h, k. Hence
∂ ∂f
∂ ∂f
(a + θ2 h, b + θ1 k) =
(a + θ3 h, b + θ4 k)
∂x ∂y
∂y ∂x
for all h and k. Taking the limit (h, k) → (0, 0) and using the assumed continuity of both
partial derivatives at (a, b) gives
∂ ∂f
∂ ∂f
(a, b) =
(a, b)
∂x ∂y
∂y ∂x
c Joel Feldman. 2014. All rights reserved.
1
January 13, 2014
The Details
1. We define F (h, k) in this way because both partial derivatives
are defined as limits of F (h, k) as h, k → 0. For example,
∂2 f
(a, b)
∂x∂y
=⇒
i
∂f
1 h ∂f
(a, b + k) −
(a, b)
h→0
k ∂x
∂x
i
1 h ∂f
∂f
lim lim F (h, k) = lim
(a, b + k) −
(a, b)
k→0 h→0
k→0 k ∂x
∂x
∂ ∂f
(a, b)
=
∂y ∂x
=⇒
i
1 h ∂f
∂f
lim F (h, k) =
(a + h, b) −
(a, b)
k→0
h ∂y
∂y
i
∂f
1 h ∂f
(a + h, b) −
(a, b)
lim lim F (h, k) = lim
h→0 k→0
h→0 h ∂y
∂y
∂ ∂f
=
(a, b)
∂x ∂y
and
∂2 f
(a, b)
∂y∂x
lim F (h, k) =
Similarly,
2. Define G(y) = f (a + h, y) − f (a, y). By the mean value theorem
1 h G(b + k) − G(b) i
F (h, k) =
h
k
1 dG
=
(b + θ1 k)
for some 0 < θ1 < 1
h dy
i
∂f
1 h ∂f
(a
+
h,
b
+
θ
k)
−
(a,
b
+
θ
k)
=
1
1
h ∂y
∂y
3. Define H(x) =
∂f
(x, b
∂y
+ θ1 k). By the mean value theorem
i
1h
H(a + h) − H(a)
h
dH
(a + θ2 h)
for some 0 < θ2 < 1
=
dx
∂ ∂f
(a + θ2 h, b + θ1 k)
=
∂x ∂y
F (h, k) =
4. Define A(x) = f (x, b + k) − f (x, b). By the mean value theorem
1 h A(a + h) − A(a) i
k
h
1 dA
(a + θ3 h)
for some 0 < θ3 < 1
=
k dx
i
∂f
1 h ∂f
(a + θ3 h, b + k) −
(a + θ3 h, b)
=
k ∂x
∂x
F (h, k) =
c Joel Feldman. 2014. All rights reserved.
2
January 13, 2014
5. Define B(y) =
∂f
(a
∂x
+ θ3 h, y). By the mean value theorem
i
1h
B(b + k) − B(b)
F (h, k) =
k
dB
(b + θ4 k)
for some 0 < θ4 < 1
=
dy
∂ ∂f
=
(a + θ3 h, b + θ4 k)
∂y ∂x
Example 2
Here is an example which shows that it is not always true that
fine
(
2
2
xy xx2 −y
if (x, y) 6= (0, 0)
+y 2
f (x, y) =
0
if (x, y) = (0, 0)
∂2 f
(a, b)
∂x∂y
=
∂2 f
(a, b).
∂y∂x
De-
This function is continuous everywhere. We now compute the first order partial derivatives.
For (x, y) 6= (0, 0)
x2 − y 2
x2 − y 2
2x
2x(x2 − y 2)
4xy 2
∂f
=
y
(x, y) = y 2
+
xy
−
xy
+
xy
∂x
x + y2
x2 + y 2
x2 + y 2
(x2 + y 2 )2
(x2 + y 2 )2
∂f
x2 − y 2
x2 − y 2
2y
2y(x2 − y 2 )
4yx2
=
x
(x, y) = x 2
−
xy
−
xy
−
xy
∂y
x + y2
x2 + y 2
x2 + y 2
(x2 + y 2)2
(x2 + y 2 )2
For (x, y) = (0, 0)
hd
i
hd i
∂f
(0, 0) =
f (x, 0)
0
=
=0
∂x
dx
dx x=0
x=0
hd
i
hd i
∂f
(0, 0) =
f (0, y)
0
=
=0
∂y
dy
dy y=0
y=0
By way of summary, the two first order partial derivatives are
( x2−y2
2 y3
y x2 +y2 + (x4x
if (x, y) 6= (0, 0)
2 +y 2 )2
fx (x, y) =
0
if (x, y) = (0, 0)
( x2 −y2
3 y2
if (x, y) 6= (0, 0)
x x2 +y2 − (x4x
2 +y 2 )2
fy (x, y) =
0
if (x, y) = (0, 0)
Both
∂f
(x, y)
∂x
and
∂f
(x, y)
∂y
are continuous. Finally, we compute
d
∂2 f
1 h2 − 02
1
(0, 0) =
fy (x, 0) x=0 = lim fy (h, 0) − fy (0, 0) = lim h 2
−
0
=1
h→0 h
h→0 h
∂x∂y
dx
h + 02
d
∂2 f
1 02 − k 2
1
(0, 0) =
fx (0, y) y=0 = lim fx (0, k) − fx (0, 0) = lim k 2
−
0
= −1
k→0 k
k→0 k
∂y∂x
dy
0 + k2
Example 2
c Joel Feldman. 2014. All rights reserved.
3
January 13, 2014