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
I think what was meant was that $\displaystyle f_{xy}= \frac{\partial^2 f}{\partial x\partial y}$. If that is what was meant, it is always true- those are different notations for the same thing.
If those really are $\displaystyle \delta$ then I would guess that they refer to "small" changes in x, y, and f and so that is an approximation to $\displaystyle f_{xy}$ 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 $\displaystyle f_{xy}= \frac{\partial^2 f}{\partial x\partial y}= f_{yx}$ which is the same as saying that $\displaystyle \frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}$- 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.
Hello, bagels0!
$\displaystyle \text{So my calculus teacher said: }\:f_{xy} \:=\: \frac{\partial^2\!f}{\partial x\,\partial y}$
$\displaystyle \text{Where exactly is this true?}$
What an UGLY way to write that identity!
Note that: .$\displaystyle f_{xy} \:=\:\frac{\partial f}{\partial y}\left(\frac{\partial f}{\partial x}\right) \:=\:\frac{\partial^2\!f}{\partial y\,\partial x}$
And so the claim is that: .$\displaystyle \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, $\displaystyle f(x,y).$