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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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2  
Derivative with respect to what? –  Javier Badia 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 $x$? –  Andres Caicedo Mar 8 '13 at 1:23
1  
Yes, what you got is correct. –  Berci Mar 8 '13 at 1:29

1 Answer 1

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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