Differentiating in two variables?

Question

So i have to find the partial derivative for both $x$ and $y$ in the equation $\frac{x^2-y^2}{x^2 + y^2}.$ I tried to use the quotient rule but the answer I get I know is not correct. I might just be making a simple arithmatic mistake but if I am I can't find it.Can someone do this with steps?

Progress

Because im finding the partial I treat the one I'm not trying to find like a constant. So I take the derivative of the numerator multiplied by the denominator, minus the derivative of the denominator mulitipled by the numerator, then take all of this and put it over the numerator squared, so I get $$\frac{ 2x(x^2+y^2)-2x(x^2-y^2) }{(x^2+y^2)^2 }$$

Answer 1

I am assuming you are having trouble simplifying, so

$$\frac{2x(x^2 + y^2) - 2x(x^2 - y^2)}{(x^2 + y^2)^2} = \frac{2xy^2 + 2xy^2}{(x^2 + y^2)^2} = \frac{4xy^2}{(x^2 + y^2)^2}.$$

Note that your answer might be slightly different. There are many forms of the same answer.

A little tip, a lot of the time you can check your answer on wolframalpha. For example, in this case you could do this.

Answer 2

First, let the equation in question be denoted by $f(x)$, ie, \begin{equation*} f(x)=\frac{x^2-y^2}{x^2+y^2}. \end{equation*} To find the partial derivative of $f$ with respect to $x$, treat the variable $y$ like a constant and apply the quotient rule. This gives us \begin{equation*} f_x=\frac{2x(x^2+y^2)-2x(x^2-y^2)}{(x^2+y^2)^2}=\frac{2xy^2+2xy^2}{(x^2+y^2)^2}=\frac{4xy^2}{(x^2+y^2)^2}. \end{equation*} For the partial derivative of $f$ with respect to $y$, treat the variable $x$ like a constant and apply the quotient rule. This gives us \begin{equation*} f_y=\frac{-2y(x^2+y^2)-2y(x^2-y^2)}{(x^2+y^2)^2}=\frac{-2yx^2-2yx^2}{(x^2+y^2)^2}=\frac{-4yx^2}{(x^2+y^2)^2}.~_{\square} \end{equation*}