TITLE: Computing a higher (2015) order partial derivative of $1/(x^2+y^2)$
QUESTION [5 upvotes]: Suppose $$f(x,y) = \frac{1}{x^2 + y^2}\text{.}$$
Find $$\frac{\partial^{2015} f}{\partial x^{2015}}\text{.}$$

REPLY [2 votes]: We have tow series, one for $x<1$ and the second for $x\geq1 $ as follow
for $x<1$
$$\frac{1}{y+x^2}=\sum_{n=0}^{\infty }(-1)^{n}\frac{x^{2n}}{y^n}
$$
$$z_x^{2015}=\sum_{n=0}^{\infty }(-1)^{n}\frac{x^{2n-2015}}{y^n}\prod_{k=0}^{2014}(2n-k)$$
For $x\geq1 $
Firstly, we will take the first derivative of $\frac{1}{y+x^2}$ which equal $\frac{-2x}{(y+x^2)^2} $
The series of $\frac{-2x}{(y+x^2)^2} $
$$z_x=\sum_{n=0}^{\infty }\frac{(-1)^{n+1}(2n+2)y^n}{x^{2n+3}}=\sum_{n=0}^{\infty }{(-1)^{n+1}(2n+2)}y^nx^{-2n-3}$$
$$z_x^{2015}=\sum_{n=0}^{\infty }{(-1)^{n+1}(2n+2)}y^nx^{-2n-3-2014}\prod_{k=0}^{2013}(-2n-3-k)$$