Equality of Mixed Partials
2
2
∂ f
∂ f
Theorem. If the partial derivatives ∂x∂y
and ∂y∂x
exist and are continuous at (x0 , y0 ),
then
∂2 f
∂2 f
(x
,
y
)
=
(x0 , y0 )
0
0
∂x∂y
∂y∂x
Proof: Here is an outline of the proof. The details are given as footnotes at the end of
the outline. Fix x0 and y0 and define(1)
F (h, k) =
1
hk
f (x0 + h, y0 + k) − f (x0 , y0 + k) − f (x0 + h, y0 ) + f (x0 , y0 )
Then, by the mean value theorem,
h
2
F (h, k) = h1 ∂f
(x0 + h, y0 + θ1 k) −
∂y
3 ∂ ∂f
(x0
∂x ∂y
=
4 1
k
h
+ θ3 h, y0 + k) −
5 ∂ ∂f
∂y ∂x (x0
+ θ3 h, y0 + θ4 k)
=
+ θ1 k)
+ θ2 h, y0 + θ1 k)
∂f
∂x (x0
F (h, k) =
∂f
(x0 , y0
∂y
∂f
∂x (x0
+ θ3 h, y0 )
i
i
for some 0 < θ1 , θ2 , θ3 , θ4 < 1. All of θ1 , θ2 , θ3 , θ4 depend on x0 , y0 , h, k. Hence
∂ ∂f
∂x ∂y (x0
+ θ2 h, y0 + θ1 k) =
∂ ∂f
∂y ∂x (x0
+ θ3 h, y0 + θ4 k)
for all h and k. Taking the limit (h, k) → (0, 0) and using the assumed continuity of both
partial derivatives at (x0 , y0 ) gives
∂ ∂f
(x0 , y0 )
∂x ∂y
=
∂ ∂f
(x0 , y0 )
∂y ∂x
The Details
(1) We define F (h, k) in this way because both partial derivatives
2
∂ f
∂y∂x (x0 , y0 )
and
are defined as limits of F (h, k) as h, k → 0. For example,
lim F (h, k) =
h→0
=⇒
1
k
h
∂f
∂x (x0 , y0
k→0 h→0
2011. All rights reserved.
h
+ k) −
1 ∂f
(x0 , y0
k→0 k ∂x
∂ ∂f
∂y ∂x (x0 , y0 )
lim lim F (h, k) = lim
=
c Joel Feldman.
∂2 f
(x0 , y0 )
∂x∂y
October 24, 2011
∂f
∂x (x0 , y0 )
+ k) −
i
∂f
(x0 , y0 )
∂x
i
Equality of Mixed Partials
1
Similarly,
lim F (h, k) =
k→0
=⇒
1
h
h
∂f
∂y (x0
=
h
1 ∂f
(x0
h→0 h ∂y
∂ ∂f
∂x ∂y (x0 , y0 )
lim lim F (h, k) = lim
h→0 k→0
+ h, y0 ) −
∂f
∂y (x0 , y0 )
+ h, y0 ) −
i
∂f
∂y (x0 , y0 )
i
(2) Define G(y) = f (x0 + h, y) − f (x0 , y). By the mean value theorem
h
i
0)
F (h, k) = h1 G(y0 +k)−G(y
k
=
=
∂f
∂y (x, y0
(3) Define H(x) =
1 dG
h dy (y0
h
1
h
∂f
(x0
∂y
+ θ1 k)
for some 0 < θ1 < 1
+ h, y0 + θ1 k) −
∂f
(x0 , y0
∂y
+ θ1 k)
i
+ θ1 k). By the mean value theorem
1
h
F (h, k) =
=
=
h
i
H(x0 + h) − H(x0 )
dH
dx (x0 + θ2 h)
∂ ∂f
∂x ∂y (x0 + θ2 h, y0
for some 0 < θ2 < 1
+ θ1 k)
(4) Define A(x) = f (x, y0 + k) − f (x, y0 ). By the mean value theorem
h
i
0)
F (h, k) = k1 A(x0 +h)−A(x
h
=
=
(5) Define B(y) =
∂f
(x0
∂x
1 dA
(x0
k dx
1
k
h
∂f
∂x (x0
+ θ3 h)
for some 0 < θ3 < 1
+ θ3 h, y0 + k) −
∂f
∂x (x0
+ θ3 h, y0 )
i
+ θ3 h, y). By the mean value theorem
F (h, k) =
=
=
1
k
h
i
B(y0 + k) − B(y0 )
dB
(y0 + θ4 k)
dy
∂ ∂f
∂y ∂x (x0 + θ3 h, y0
for some 0 < θ4 < 1
+ θ4 k)
Example. Here is an example which shows that it is not always true that
2
∂ f
∂y∂x (x0 , y0 ).
∂2 f
(x0 , y0 )
∂x∂y
=
Define
f (x, y) =
c Joel Feldman.
2011. All rights reserved.
2
2
xy xx2 −y
+y 2
0
if (x, y) 6= (0, 0)
if (x, y) = (0, 0)
October 24, 2011
Equality of Mixed Partials
2
This function is continuous everywhere. We now compute the first order partial derivatives.
For (x, y) 6= (0, 0)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
∂f
∂x (x, y)
2x(x −y )
x −y
4xy
2x
= y xx2 −y
+y 2 + xy x2 +y 2 − xy (x2 +y 2 )2 = y x2 +y 2 + xy (x2 +y 2 )2
∂f
∂y (x, y)
2y(x −y )
−y
2y
4yx
= x xx2 +y
= x xx2 −y
2 − xy x2 +y 2 − xy
+y 2 − xy (x2 +y 2 )2
(x2 +y 2 )2
For (x, y) = (0, 0)
∂f
∂x (0, 0)
∂f
(0, 0)
∂y
d =
f
(x,
0)
dx
dx 0 x=0 = 0
x=0
d
d = dy f (0, y) y=0 = dy 0 y=0 = 0
=
d
By way of summary, the two first order partial derivatives are
Both
∂f
(x, y)
∂x
∂2 f
(0, 0)
∂x∂y
and
fx (x, y) =
(
+
y xx2 −y
+y 2
0
fy (x, y) =
(
x xx2 −y
+y 2 −
0
∂f
(x, y)
∂y
2
2
4x2 y 3
(x2 +y 2 )2
if (x, y) 6= (0, 0)
if (x, y) = (0, 0)
2
2
4x3 y 2
(x2 +y 2 )2
if (x, y) 6= (0, 0)
if (x, y) = (0, 0)
are continuous. Finally, we compute
h2 −02
1
1
f
(x,
0)
=
lim
f
(h,
0)
−
f
(0,
0)
=
lim
h
−
0
=1
y
y
y
2
2
dx
h +0
x=0
h→0 h
h→0 h
d
02 −k2
∂2 f
1
1
∂y∂x (0, 0) = dy fx (0, y) y=0 = lim k fx (0, k) − fx (0, 0) = lim k k 02 +k2 − 0 = −1
=
d
k→0
c Joel Feldman.
2011. All rights reserved.
k→0
October 24, 2011
Equality of Mixed Partials
3