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I want to solve $\displaystyle f_1(x,y)$ where $f(x,y)= \dfrac{x}{x^2+y^2}$ by hand.

What steps are involved?

Update: $\displaystyle f_1(x,y)$ denotes $\dfrac{\partial}{\partial x}f(x,y)$

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    with respect to $x$, $y$ or some other variable?2012-09-27
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    Updated question.2012-09-27
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    What exactly do you want? Is it to find a function $f(x,y)$ such that ${\partial\over\partial x} f(x,y)={x\over x^2+y^2}$?2012-09-27
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    @DavidMitra Why the down vote? I want do derive with regards to x and have successfully done so with the help of the quotient rule.2012-09-27
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    I didn't downvote. But I suspect whoever did it, did so because your question is not all all clear. If you wish to compute a partial derivative, you should ask: "Let $f(x,y)={x\over x^2+y^2}$. How do I find $f_1(x,y)$, where $f_1$ denotes ..." Do *not* assume people know what your notation means. The way your question is currently phrased led me to my previous comment.2012-09-27
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    @DavidMitra Ok thanks for the comment, I will be careful :). I just assumed it was standard as the school book used it: http://www.amazon.com/Calculus-Complete-Course-Seventh-7th/dp/0321549287/, He says f1 denotes derivation with regards to the first parameter and f2 with regards to the second and so on.2012-09-27
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    So, you need to edit your question. $f(x,y)={x\over x^2+ y^2}$ (*not $f_1$*). So say "Let $f(x,y)={x\over x^2+y^2}$. I want to find $f_1(x,y)$, where $f_1(x,y)=$... ", or "I want to find $f_1(x,y)$ where $f(x,y)={x\over x^2+y^2}$. Update ...".2012-09-27
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    @DavidMitra thanks again updated my question2012-09-27

2 Answers 2

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You use the quotient rule to get $\left(\frac x{x^2+y^2}\right)'=\frac {x'(x^2+y^2)-(x^2+y^2)'x}{(x^2+y^2)^2}$ then take the indicated derivatives in the numerator.

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    Thanks I solved it beautifully with the quotient rule.2012-09-27
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Treat $y$ like a constant and use quotient rule to derive with respect to $x$. Same idea for deriving with respect to $y$.