# Calculate `(df)/(dx)` and `(df)/(dy)` in the image attached.

*print*Print*list*Cite

### 1 Answer

You need to evaluate partial derivative `(del f)/(del x)` differentiating the function with respect to x, considering y as constant, such that:

`(del f)/(del x) = (del (x^2y/(x^2+y^2)))/(del x)`

Since the function contains the variable x at numerator and denominator, you need to differentiate using quotient rule, such that:

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

`(del (x^2y/(x^2+y^2)))/(del x) = (2xy(x^2+y^2) - x^2y*(2x))/((x^2+y^2)^2)`

Factoring out `2xy` to numerator yields:

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

Reducing duplicate terms yields:

`(del (x^2y/(x^2+y^2)))/(del x) = (2xy^3)/((x^2+y^2)^2)`

You need to evaluate partial derivative `(del f)/(del y)` differentiating the function with respect to y, considering x as constant, such that:

`(del f)/(del y) = (del (x^2y/(x^2+y^2)))/(del y)`

`(del f)/(del y) = (x^2(x^2+y^2) - x^2y*2y)/((x^2+y^2)^2)`

Factoring out `x^2` yields:

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

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

**Hence, evaluating the partial derivatives, yields `(del f)/(del x) = (2xy^3)/((x^2+y^2)^2)` and **`(del f)/(del y) = (x^2(x^2 - y^2))/((x^2+y^2)^2).`