# What steps are involved to derive $\displaystyle f_1(x,y)$ where $f(x,y)= \frac{x}{x^2+y^2}$

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? –  enzotib Sep 27 '12 at 16:31
Updated question. –  Farmor Sep 27 '12 at 16:41
What exactly do you want? Is it to find a function $f(x,y)$ such that ${\partial\over\partial x} f(x,y)={x\over x^2+y^2}$? –  David Mitra Sep 27 '12 at 16:43
@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. –  Farmor Sep 27 '12 at 16:47
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. –  David Mitra Sep 27 '12 at 16:51
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.
Treat $y$ like a constant and use quotient rule to derive with respect to $x$. Same idea for deriving with respect to $y$.