View source for Failure of Clairaut's theorem where only one of the mixed partials is defined

==Statement==

===For a function of two variables at a point===

It is possible to have a function <math>f</math> of two variables <math>x,y</math> and a point <math>(x_0,y_0)</math> in the [[domain]] of <math>f</math> such that the [[second-order mixed partial derivative]] <math>f_{xy}(x_0,y_0)</math> exists but the second-order mixed partial derivative <math>f_{yx}(x_0,y_0)</math> does not exist.

==Proof==

===Example===

Consider a function defined on all of <math>\R^2</math> as:

<math>f(x,y) := \left\lbrace \begin{array}{rl} 0, & y = 0 \\ 1, & y \ne 0 \\\end{array} \right.</math>

We note that:

* Since <math>f</math> depends only on <math>y</math>, <math>f_x(x,y)</math> is identically the zero function.
* Thus, the second-order mixed partial derivative <math>f_{xy}(x,y)</math> is identically the zero function.
* On the other hand, <math>f_y(x,y) = 0</math> for <math>y \ne 0</math> and <math>f_y(x,y)</math> is undefined for <math>y = 0</math>.
* In particular, this means that <math>f_{yx}(x,y)</math> is not defined on the line <math>y = 0</math> (i.e., the <math>x</math>-axis).