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Prove $$F_{xy}(x,y)F(x,y) = F_x(x,y)F_y(x,y)$$

$F(x,y)$ is separable.

This is such a wierd question, ... maybe its just me ... how do I start?


What I tried anyways:

$F(x,y)=f(x)g(y)$

$F_x(x,y) = \frac{d}{dx} f(x) g(y)$

$F_y(x,y) = \frac{d}{dy} f(x) g(y)$

$F_{xy}(x,y) = \frac{d}{dx} (\frac{d}{dy} (f(x)g(y)))$

None of the equations seem to give me any ideas :(

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1 Answer 1

up vote 2 down vote accepted

If $F(x,y)=f(x) g(y)$ and the required derivatives exist, just compute the required partials.

Note these are partial derivatives. When finding $F_x(x,y)$, for example, you think of the variable $y$ as fixed and differentiate with respect to $x$: $$ F_x(x,y) ={\textstyle{\partial \over\partial x}}\bigl[\, f(x)g(y)\,\bigr] = g(y) {\textstyle{\partial \over\partial x}} f(x) =g(y){\textstyle{d\over dx} }f(x)= g(y)f'(x). $$

You have:

$\ \ \ \ F_x(x,y)=f'(x) g(y)$,

$\ \ \ \ F_y(x,y)=f(x)g'(y)$,

and

$\ \ \ \ F_{xy}(x,y)={\partial\over\partial y}F_x(x,y) ={\partial\over\partial y} \bigl[\,f'(x)g(y)\,\bigr] =f'(x)g'(y).$

Now substitute into your equation and show that it's true.


Remark: Note you reversed the order of differentiation in your expression for $F_{xy}=(F_x)_y$; you find $F_x$ first. (It is not always the case that $F_{xy}=F_{yx}$.)

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:( Its so simple, yet I can't see it ... I guess the key is "think of the variable $y$ as fixed and differentiate with respect to $x$" –  Jiew Meng Apr 11 '12 at 14:05

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