# Where is this statement true?

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• Oct 20th 2011, 03:07 PM
bagels0
Where is this statement true?
So, my calculus teacher just said in class today that, fxy = ((δ^2)f)/(δxδy) is true.

Where exactly is this true?

Thanks.
• Oct 20th 2011, 03:13 PM
TheChaz
Re: Where is this statement true?
Quote:

Originally Posted by bagels0
So, my calculus teacher just said in class today that, fxy = ((δ^2)f)/(δxδy) is true.

Where exactly is this true?

Thanks.

Not sure exactly what you're saying, but it brings this to mind.
Symmetry of second derivatives - Wikipedia, the free encyclopedia
• Oct 21st 2011, 10:29 AM
HallsofIvy
Re: Where is this statement true?
I think what was meant was that $f_{xy}= \frac{\partial^2 f}{\partial x\partial y}$. If that is what was meant, it is always true- those are different notations for the same thing.

If those really are $\delta$ then I would guess that they refer to "small" changes in x, y, and f and so that is an approximation to $f_{xy}$ but is exact in the case that f is a linear function of x and y.

A third possibility is that your teacher was saying that $f_{xy}= \frac{\partial^2 f}{\partial x\partial y}= f_{yx}$ which is the same as saying that $\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}$- that is that the "mixed" second derivative are equal- independent of the order of differentiation. That is true as long as the second partial derivatives are continuous.
• Oct 21st 2011, 12:42 PM
Soroban
Re: Where is this statement true?
Hello, bagels0!

Quote:

$\text{So my calculus teacher said: }\:f_{xy} \:=\: \frac{\partial^2\!f}{\partial x\,\partial y}$

$\text{Where exactly is this true?}$

What an UGLY way to write that identity!

Note that: . $f_{xy} \:=\:\frac{\partial f}{\partial y}\left(\frac{\partial f}{\partial x}\right) \:=\:\frac{\partial^2\!f}{\partial y\,\partial x}$

And so the claim is that: . $\begin{Bmatrix} \dfrac{\partial^2\!f}{\partial y\,\partial x} \:=\: \dfrac{\partial^2\!f}{\partial x\,\partial y} \\ \text{or} \\ f_{xy} \:=\:f_{yx} \end{Bmatrix}$

This is true for all functions, $f(x,y).$