Counterexample for Mixed Partial Derivative
For most textbook examples in multivariate calculus fxy(a, b) = fyx(a, b). The following counterexample shows
a function where fx and fy are continuous everywhere and yet fxy(0, 0) ≠ fyx(0, 0) because they are not
continuous functions at the origin. In fact, fxy(0, 0) = 1 and fyx(0, 0) = -1.
Clairaut’s Theorem Suppose f is defined on a disk D that contains the point (a, b). If the functions fxy and fyx
are both continuous on D, then fxy(a, b) = fyx(a, b).
 x 2  y2
xy
if ( x, y)  (0,0)
f ( x , y)   x 2  y 2

if ( x, y)  (0,0)
0
f(x, y) is continuous
r 2 (cos 2   sin 2 ) r 2 sin 4
Using polar coordinates, x  r cos  and y  r sin  , f(x, y) = f (r, )  r cos  sin  2

4
r (cos 2   sin 2 )
2
Therefore |f(xy)| ≤ r2/4 → 0 as r → 0, and f(x, y) is continuous at the origin.
fx(x, y) and fy(x, y) are both continuous
f x ( x , y) 
x 4 y  4x 2 y 3  y 5
 r (cos 4  sin   4 cos 2  sin 3   sin 5 )  6r  0 as r  0
(x 2  y 2 ) 2
h2  0
(h )(0) 2
0
f (0  h,0)  f (0,0)
0
h 0
f x (0,0)  lim
 lim
 lim  0
h 0
h 0
h 0 h
h
h
f y ( x , y) 
x 5  4x 3 y 2  xy 4
 r (cos 5   4 cos 3  sin 2   cos  sin 4 )  6r  0 as r  0
2
2 2
(x  y )
f y (0,0)  lim
k 0
f (0,0  k )  f (0,0)
 lim
k 0
k
(0)(k )
0  k2
0
0
0  k2
 lim  0
k

0
k
k
fxy(x,y) and fyx(x,y) are not continuous at the origin
f xy ( x, y)  f yx ( x, y) 
x 6  9x 4 y 2  9x 2 y 4  y 6
 cos 6   9 cos 4  sin 2   9 cos 2  sin 4   sin 6   cos 2(1  2 sin 2 2)
2
2 3
(x  y )
If θ = 0 (x-axis), then fxy(x,y) = 1. If θ = π/2 (y-axis), then fxy(x,y) = -1. fxy(x, y) is dependent on θ only.
 k5
0
4
f (0,0  k )  f x (0,0)
f xy (0,0)  lim x
 lim k
 lim (1)  1
k 0
k 0
k 0
k
k
h5
0
4
f y (0  h,0)  f y (0,0)
h
f yx (0,0)  lim
 lim
 lim (1)  1
h 0
h 0
h 0
h
h