21.6. MIXED PARTIAL DERIVATIVES 455

Note that (x+ t,y+ s) ∈U because

|(x+ t,y+ s)− (x,y)|= |(t,s)|=(t2 + s2)1/2 ≤

(r2

4+

r2

4

)1/2

=r√2< r.

As implied above, h(t) ≡ f (x+ t,y+ s)− f (x+ t,y). Therefore, by the mean value theo-rem from calculus and the (one variable) chain rule,

∆(s, t) =1st(h(t)−h(0)) =

1st

h′ (αt) t =1s( fx (x+αt,y+ s)− fx (x+αt,y))

for some α ∈ (0,1). Applying the mean value theorem again, ∆(s, t) = fxy (x+αt,y+β s)where α,β ∈ (0,1).

If the terms f (x+ t,y) and f (x,y+ s) are interchanged in 21.6, ∆(s, t) is also un-changed and the above argument shows there exist positive numbers γ,δ ∈ (0,1) suchthat ∆(s, t) = fyx (x+ γt,y+δ s) . Letting (s, t)→ (0,0) and using the continuity of fxy andfyx at (x,y) , lim(s,t)→(0,0) ∆(s, t) = fxy (x,y) = fyx (x,y) .

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

Corollary 21.6.2 Suppose U is an open subset of Rn and f : U → R has the propertythat for two indices k, l, fxk , fxl , fxlxk , and fxkxl exist on U and fxkxl and fxlxk are bothcontinuous at x ∈U. Then fxkxl (x) = fxlxk (x).

It is necessary to assume the mixed partial derivatives are continuous in order to assertthey are equal. The following is a well known example [3].

Example 21.6.3 Let f (x,y) =

{xy(x2−y2)

x2+y2 if (x,y) ̸= (0,0)0 if (x,y) = (0,0)

.

Here is a picture of the graph of this function. It looks innocuous but isn’t.

From the definition of partial derivatives it follows immediately thatfx (0,0) = fy (0,0) = 0. Using the standard rules of differentiation, for (x,y) ̸= (0,0),

fx = yx4 − y4 +4x2y2

(x2 + y2)2 , fy = xx4 − y4 −4x2y2

(x2 + y2)2

Now fxy (0,0)≡ limy→0fx(0,y)− fx(0,0)

y = limy→0−y4

(y2)2 =−1 while the mixed partial deriva-

tive in the other order is fyx (0,0)≡ limx→0fy(x,0)− fy(0,0)

x = limx→0x4

(x2)2 = 1 showing that,

although the mixed partial derivatives do exist at (0,0), they are not equal there.