# Find both first partial derivatives.?f(x,y)=xy/(x^2+y^2) f x(x,y)= f y(x,y)=

### 1 Answer | Add Yours

`f(x,y)=(xy)/(x^2+y^2)`

To determine `f_x(x,y)` , take the derivative of the function with respect to x using quotient rule. And treat y as constant.

`f_x(x,y) = ((x^2+y^2) (xy)' - (xy)(x^2+y^2)')/(x^2+y^2)^2`

`f_x(x,y) = ((x^2+y^2)(y) -(xy)(2x))/(x^2+y^2)^2`

To simplify the numerator, factor out the GCF which is y.

`f_x(x,y)=(y(x^2+y^2-x(2x)))/(x^2+y^2)^2`

`f_x(x,y)=(y(x^2+y^2-2x^2))/(x^2+y^2)^2`

So,

`f_x(x,y)=(y(y^2-x^2))/(x^2+y^2)^2`

To solve for `f_y(x,y)` , take the derivative of the function with respect to y. Apply quotient rule again. But, it is the x that is constant.

`f_y(x,y)= ((x^2+y^2)(xy)' - (xy)(x^2+y^2))/(x^2+y^2)^2` ` `

`f_y(x,y)=((x^2+y^2)(x)-(xy)(2y))/(x^2+y^2)^2`

To simplify the numerator, factor out the GCF which is x.

`f_y(x,y)=(x(x^2+y^2-y(2y)))/(x^2+y^2)^2`

`f_y(x,y)=(x(x^2-y^2-2y^2))/(x^2+y^2)^2`

So,

`f_y(x,y)=(x(x^2-y^2))/(x^2+y^2)^2`

==============================================

**Hence, the first order partial derivatives of the function are:**

**`f_x(x,y)=(y(y^2-x^2))/(x^2+y^2)^2 ` and ****`f_y(x,y)=(x(x^2-y^2))/(x^2+y^2)^2` .**