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In math, for most functions of $x$ and $y$, $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}$$ That is, the partial derivative of $f(x, y)$ with respect to $x$ then $y$ is the same as the partial derivative with respect to $y$, then $x$.

I have heard that there are functions for which this is not true. Can someone give me some examples?

Sebastiano
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Andrew Brown
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  • https://math.stackexchange.com/questions/219759/show-that-both-mixed-partial-derivatives-exist-at-the-origin-but-are-not-equal?rq=1 – Fred Jul 21 '21 at 11:55
  • You want something with a slope in one direction but no well defined slope in the other. Take a balloon and pinch the top between your fingers. The balloon has a slope parallel to the pinch but not across the pinch. – Paul Jul 21 '21 at 12:01
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    Such a function must have $f_{xy}$ or $f_{yx}$ not continuous: https://calculus.subwiki.org/wiki/Clairaut%27s_theorem_on_equality_of_mixed_partials. – preferred_anon Jul 21 '21 at 12:02

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On Wikipedia you can see the canonical example of the Italian mathematicians, Giuseppe Peano.

Here there is his example of a function with mixed partial second derivatives, it must have such non-continuous derivatives as in the following example Given a continuous function: $$f(x,y)=\left\{\begin{matrix} xy \dfrac{x^2-y^2}{x^2+y^2} & \forall (x,y) \in \Bbb R^2 \setminus \{(0,0)\}\\ 0 &(x,y) = (0,0) \end{matrix}\right.$$

We have continuous prime partial derivatives: $$f_x (x,y) =\left\{\begin{matrix} y \dfrac{x^2-y^2}{x^2+y^2} + xy \frac {2x (x^2+y^2) - 2x (x^2-y^2)}{(x^2+y^2)^2} & \forall (x,y) \in \Bbb R^2 \setminus \{(0,0)\}\\0 & (x,y) = (0,0) \end{matrix}\right.$$

$$f_y (x,y)=\left\{\begin{matrix} - x {y^2-x^2}{x^2+y^2} - xy \dfrac{2y (x^2+y^2) - 2y (y^2-x^2)}{(x^2+y^2)^2} & \forall (x,y) \in \Bbb R^2 \setminus \{(0,0)\} \\0 & (x,y) = (0,0) \end{matrix}\right.$$ But the mixed second derivatives are not continuous and are different, in fact: $$f_{xy} (0,0) = \lim_{k \to 0} \frac{f_x (0,k) - f_x (0,0)}{k} = -1, \quad f_{yx} (0,0) = \lim_{h \to 0} \frac{f_y (h,0) - f_y (0,0)}{h} = +1$$So $$f_{yx} \neq f_{xy}$$

Here $f_x=\frac{\partial f}{\partial x}$ and $f_y=\frac{\partial f}{\partial y}$, $f_{xx}=\frac{\partial^2 f}{\partial x^2}$ and $f_{yx}=\frac{\partial^2 f}{\partial y\partial x} $.

Sebastiano
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