Under certain conditions themixed partial derivatives will always be equal. The simple condition is
that if they exist and are continuous, then they are equal. This astonishing fact is due to Euler in 1734 and
was proved by Clairaut although not very well. The first satisfactory proof was by Hermann Schwarz in
1873. For reasons I cannot understand, calculus books hardly ever include a proof of this important
result. It is not all that hard. It is based on the mean value theorem for derivatives. Here it
is.

Theorem 3.5.1Suppose f : U ⊆ ℝ^{2}→ ℝ where U is an open set on which f_{x},f_{y}, f_{xy}and f_{yx}exist.Then if f_{xy}and f_{yx}are continuous at the point

(x,y)

∈ U, it follows

fxy(x,y) = fyx(x,y).

Proof:Since U is open, there exists r > 0 such that B

((x,y),r)

⊆ U. Now let

|t|

,

|s|

< r∕2 and
consider

---------h(t)--------- -------h(0)-------
Δ (s,t) ≡ 1-{◜f (x+ t,y+ s◞◟)− f (x + t,y◝) −◜(f (x,y + ◞s◟)− f (x,y◝))}. (3.6)
st

is also unchanged and the above
argument shows there exist γ,δ ∈

(0,1)

such that

Δ (s,t) = fyx(x+ γt,y+ δs).

Letting

(s,t)

→

(0,0)

and using the continuity of f_{xy} and f_{yx} at

(x,y)

,

(s,lt)im→(0,0)Δ (s,t) = fxy(x,y) = fyx(x,y).

This proves the theorem.

The following is obtained from the above by simply fixing all the variables except for the two of
interest.

Corollary 3.5.2Suppose U is an open subset of ℝ^{n}and f : U → ℝ has the property that for twoindices, k,l, f_{xk}, f_{xl},f_{xlxk}, and f_{xkxl}exist on U and f_{xkxl}and f_{xlxk}are both continuous at x ∈ U.Then f_{xkxl}

(x)

= f_{xlxk}

(x )

.

It is necessary to assume the mixed partial derivatives are continuous in order to assert they are equal.
The following is a well known example [6].

Example 3.5.3Let

{
xy(x2−y2)if (x,y) ⁄= (0,0)
f (x,y) = x2+y2
0 if (x,y) = (0,0)

From the definition of partial derivatives it follows immediately that f_{x}

(0,0)

= f_{y}

(0,0)

= 0. Using the
standard rules of differentiation, for

(x,y)

≠

(0,0)

,

f = yx4-− y4-+-4x2y2,f = xx4-− y4-−-4x2y2
x (x2 + y2)2 y (x2 + y2)2

Now

f (0,0) ≡ lim fx-(0,y)−-fx(0,0)
xy y→0 y
− y4
= lyim→0(y2)2 = − 1