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$ d=\frac{x^2}{x-y}$ and im trying to find d'.

do i take it as $x^2(x-y)^{-1}$ ? That seems to give me the wrong answer.

If i take it like that, i get $ 2x(x-y)^{-1} + x^2(-1)(x-y)^{-2}$.

is this correct?

with respect to $x$

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Derivative with respect to what? – Javier Mar 8 '13 at 1:19
You want to find the derivative of $$\frac{x^2}{x-y}$$ with respect to $x$? – Pedro Tamaroff Mar 8 '13 at 1:19
yes sorry with respect to x – user65678 Mar 8 '13 at 1:20
Is $y$ a different variable? A function of $x$? – Andrés E. Caicedo Mar 8 '13 at 1:23
Yes, what you got is correct. – Berci Mar 8 '13 at 1:29

If we have $f(x,y)=\frac{x^2}{x-y}$ and we are only differentiating with respect to $x$ we can use either the product rule, or the quotient rule. I prefer the product rule though. So here's what we do :

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

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thank you. im on the right track it seems. guess i just messed up somewhere on my way to the final answer in my problem. – user65678 Mar 8 '13 at 1:31
@Calc1DropOut what prime means for the function of two variables? – Kaster Mar 8 '13 at 1:32
@Kaster I edited to make it more clear. – MITjanitor Mar 8 '13 at 1:36
@Calc1DropOut I bet you meant $\frac {\partial f}{\partial x}$ :) – Kaster Mar 8 '13 at 1:38
Yeah ;) thanks! – MITjanitor Mar 8 '13 at 1:41

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