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Can you please explain how to differentiate $f(x,y)$ with respect to $x-y$ ?

Thank you very much.

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Are you familiar with the notion of a directional derivative? –  Jonathan Christensen Dec 19 '12 at 18:45
    
No, I do not. I started looking at it, how can I use it to address this question? –  Essa Dec 19 '12 at 23:11
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2 Answers

For a general co-ordinate $u$, $$\frac{df}{du} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}$$

So for $u = x-y$, we have:

$$\frac{df}{du} = \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}$$

Note that the derivative given above is different to the partial derivative $\frac{\partial f}{\partial u}$ which will in general depend on a given parameterisation of the curve as $f = f(u,v)$, as mentioned in another answer.

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Thank you, can you please say what determines the parameterization of the curve? I am new to this subject. The function is of the form f(x,y) = (a-b x)^2 - (c - d y)^2 and the question is how does f change when (x-y) change? –  Essa Dec 19 '12 at 19:25
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Depends on what the other variable in your coordinate system is. For example, if your new system is $u = x - y$, $v = x$ and $f(x, y) = x + y$, then $$ f(u, v) = -u + 2v $$ and so the derivative with respect to $u$ is $-1$. But if the new system is $u = x - y$, $v = y$, then $$ f(u, v) = u + 2v $$ and the derivative is $1$.

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Thank you. I am not sure what do you mean by the other variable. What determines how to define the new system? The function is of the form f(x,y) = (a-b x)^2 - (c - d y)^2 and the question is how does f change when (x-y) change? –  Essa Dec 19 '12 at 19:23
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