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As i know from the text book $\frac{d}{dx}$$F(x,y)$=$F_1 (x,y)$ but i wonder why it is not equal to$F_1(x,y)+F_2 (x,y)$$\frac{dy}{dx}$

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What? Please give the definition of differentiation you're working with. –  Alex Becker Jan 2 '12 at 8:29
    
$F$ is just any differentiable function with 2 variable x and y and the $F_1$ is the partial derivative –  johnny Jan 2 '12 at 8:35
    
The partial derivative with respect to x? This is usually written $F_x$. –  Alex Becker Jan 2 '12 at 8:36
    
really?my text book usually use $F_1$. –  johnny Jan 2 '12 at 8:39
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In any case if you mean by F_ the partial derivative to y, then $dy/dx$ is usually zero unles the variabele $x$ would be depended of $x$ as well. Usually this is not the case in these exercices. –  MrOperator Jan 2 '12 at 8:41
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closed as not a real question by Austin Mohr, tomasz, William, Quixotic, rschwieb Sep 14 '12 at 16:36

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1 Answer

It seems there is confusion between the partial derivative, and the full derivative. Indeed, the expression you wrote is correct - the full derivative is of F(x,y) in respect to x is: $$\frac{d}{dx}F(x,y)=F_x (x,y) + \frac{dy}{dx} F_y (x,y)$$ The partial derivative $\frac{\partial}{\partial x} F(x,y) = F_x (x, y)$ is just an expression which calculates the function's dependence on x, while assuming that x is independent from y. When having concrete x and y this may not be the case, and the previous expression - which is a simple application of the chain rule - is applicable.

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