# Second order partial derivatives - notation

I have seen both of these used, and people around me seem to disagree, so which one is correct: (first derivative with respect to x, then y):

(1) $$\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}) = \frac{\partial^{2} f}{\partial x\partial y}$$

(2) $$\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}) = \frac{\partial^{2} f}{\partial y\partial x}$$

and why? (reasons, history?)

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$\def\part#1#2{{\partial#1\over\partial#2}}$ $\def\parts#1#2#3{{\partial^2#1\over\partial#2\,\partial#3}}$

On the left hand side of your equations, you have the symbol "$\part{\vphantom f}y\bigl(\part f x\bigr)"$. By definition this is the partial derivative of the function $\part fx$ with respect to $y$. So, upon encountering this symbol, you take the function $\part fx$ and then take its partial with respect to $y$. The natural notation of the type on the right hand side of your equations is the notation used in (2) of your post: $$\tag{3} \part{\vphantom f}y\Bigl(\part f x\Bigr)=\parts f y x.$$

I will not surmise why this is the "natural" notation, but will point out that $(3)$ gives the adopted definition for the symbol $\parts f y x$ in any calculus/analysis text, or any other "credible" source, you'll find.

I emphasise here that $(3)$ defines the symbol $\parts f y x$; that this sometimes gives an expression that equals $\parts f x y$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.)

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Ok, so I think that this was the answer that I was looking for. I had seen the other notation in a business calculus book and they were being pretty consistent about it. I had also asked a Ph.D. student (with specialty in analysis) and he said that he used the first (1) notation. Can you think of any reason why someone might want the other notation? (option (1)) –  Thomas Apr 8 '12 at 23:21
@Thomas Most, if not all, texts (including Paul's online notes, Stewart's Calculus, Apostol's Calculus, etc...) I've seen use (2) as the definition. It does seem natural to me since you can "multiply the left side" to get the right side. I can think of no reason to prefer the other. In any case, the symbol does need a definition, as the two expressions can give different results for not so nice functions. For example, there are differentiable functions with unequal mixed derivatives. –  David Mitra Apr 8 '12 at 23:35

The order is important when the function is not $C^2$. That is, the second derivatives (in relation to any combination of two variables) of $f$ are continuous functions. If the function is $C^2$ then it doesn't matter the order in which the variables appear.

This is a widely known result called Schwarz's Theorem, but it seems that there are other names for it. Check out for more in http://en.wikipedia.org/wiki/Symmetry_of_partial_derivatives under the "Clairaut's theorem" subtitle.

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please explain all difference between all the cases and how come they are different? –  Victor Apr 2 '12 at 20:04
@Victor Have you taken a multi variable calculus course? If so, it should be clear that they will infact be different for some classes of functions and whatnot! –  user21436 Apr 2 '12 at 20:08
@KannappanSampath - no, never taken it before. –  Victor Apr 2 '12 at 20:10
@Victor Then I suggest you do it. :P –  user21436 Apr 2 '12 at 20:10
Hi, thanks for the response, but I was asking about which notation is correct. –  Thomas Apr 2 '12 at 20:17

here i uesd notation $f_{xy}$ and $f_{yx}$ for double derivative.\ There is some theorem on equality of $f_{xy}$ and $f_{yx}$. Theorem : (Young's theorem)\ Let $f$ be a real function defined on non-empty open set $E$ subset of $R^{2}$. If $f_{x}$ and $f_{y}$ exist in some nbhd. of $(x,y)$ and are both differentiable at point $(x,y)$ with respect to $x$ and $y$ then $f_{xy}=f_{yx}$ at point $(x,y)$.

Theorem: (Schwartz's theorem) Let $f:E\to R$ be a function such that its partial derivatives $f_{x},f_{y}$ and $f_{xy}$ exist and are continuous in a nbhd. of a point $(x,y)$ then $f_{yx}$ exist such that $f_{xy}=f_{yx}$.

Note: One can see that the converse of above theorem need not be true.

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I always saw the second : the last operation you made is derivating with respect Y and tha symbol appears in the same order . In the other notation the order is inverse fxy it means the last operation you made is derivating with respect y. But : if both fxy and fyx are defined at a neighborhood of a point and are continuous at the point their value is the same at the point.See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10

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See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10 –  alpha.Debi Apr 8 '12 at 11:01