# please find the partial derivatives fx and fy (1) f(x,y)= x^2-y^2/x-yhelp

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### 2 Answers

You need to find the first order partial derivative `f_x(x,y), ` hence, you need to differentiate the given function with respect to x, considering y as constant such that:

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

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

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

`f_x(x,y) = (4x^2 + 2xy - 2x^2 + 2y^2)/((2x+y)^2) `

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

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

You need to find the first order partial derivative `f_y(x,y), ` hence, you need to differentiate the given function with respect to y, considering x as constant such that:

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

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

`f_y(x,y) = (-4xy - 2y^2 - x^2 + y^2)/((2x+y)^2)`

`f_y(x,y) = -(4xy + y^2 + x^2)/((2x+y)^2)`

**Hence, evaluating the first order partial derivatives yields `f_x(x,y) = (2(x^2 + xy + y^2))/((2x+y)^2)` and `f_y(x,y) = -(4xy + y^2 + x^2)/((2x+y)^2).` **

### User Comments

sorry I made a mistake in my question.... it should be divided by 2x + y ...and not x-y

so it is really ... f(x,y) = x^2-y^2/2x+y

I hope that i can be helped