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When taking double/triple derivatives of multivariable functions, we see that no matter which order we take the derivative in, as long as we take the derivative od the function with respect to the same number of variables, the same number of times, our answer is the same? Is there an intuitive explanation for this? I cannot seem to understand how.

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    Your statement is not true in general. – Mark Viola Oct 31 '19 at 02:29
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    They aren’t always the same. Google “Clairaut’s Theorem”. – MPW Oct 31 '19 at 02:29
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    This is true when the function has continuous second partial derivatives, but not in general. – Math1000 Oct 31 '19 at 03:38
  • Please do some basic research. They aren't always the same. –  Oct 31 '19 at 06:43
  • For the reason why this is true for sufficiently regular functions, there are some good answers here: https://math.stackexchange.com/questions/170764/ – Ted Oct 31 '19 at 06:56
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    higher order partial derivatives of smooth functions (or sufficiently smooth) are equal regardless of the order one takes them. This is the Schwarz theorem. – AlvinL Oct 31 '19 at 08:35

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I agree with MPW and decided to give an example of his claim:

Consider the function: $f:\Bbb R^2\to \Bbb R$ defined by $$f(x,y)=\begin{cases}\frac{xy(x^2-y^2)}{x^2+y^2}\text{ for $(x,y)\ne (0,0)$}\\0\text{ for $(x,y)=(0,0).$}\end{cases}$$

The surface looks like: Graph of the function

Now if we calculate $\displaystyle \frac{\partial^2 }{\partial x\partial y}f(x,y)|_{(0,0)}$ and $\displaystyle \frac{\partial^2 }{\partial y\partial x}f(x,y)|_{(0,0)}$ we get $\displaystyle \frac{\partial^2 }{\partial x\partial y}f(x,y)|_{(0,0)}=1$ and $\displaystyle \frac{\partial^2 }{\partial y\partial x}f(x,y)|_{(0,0)}=-1$. So the commutativity of the second order partial derivatives does not work always.

For a detailing here is a link : Symmetry of the Second derivative.

Hope it works.

Sujit Bhattacharyya
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They are not always equal. What is in fact true is that if, for example, you let $u$ be a function of $(x,y),$ and $u_x,u_y,u_{xy}$ all exist and $u_{xy}$ is continuous, then it follows that $u_{yx}$ also exists and is equal to $u_{xy}.$ (Of course, we can replace $u_{xy}$ with $u_{yx}$ everywhere above without altering the truth of the statement -- the point is that if one of the second mixed partials is continuous, then the other exists and is the same as the first).

So why is this true? Well, you ask for an intuitive understanding -- it's notoriously difficult to pin down what that means, but let's try. In this case it's even harder because one is hard pressed to find an intuitive understanding of a mixed derivatives. Say you think in terms of slope, then the second derivative is the rate of change of slope -- but the problem here is that the second derivative is in a direction completely orthogonal to the first. I think the best best here would be to understand as many proofs of this result as possible, in order to see this from different perspectives. Then perhaps you'll begin to form your own understanding of this fact.

Allawonder
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