Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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

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

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

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

share|cite|improve this question
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:

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

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


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

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 $F_{xy}=F_{yx}$.)

share|cite|improve this answer
:( 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.