Finding the $2n$ th derivative of $\frac{1}{1+x^2y^2}$ with repect to $x$

Question

We know that with repect to $x$,

\begin{align} \left( \frac{1}{1-x^2 y^2} \right)^{(2n)} = \frac{(2n)!y^{2n}}{2 } \left( \frac{1}{(1+xy)^{2n+1}} +\frac{1}{(1-xy)^{2n+1}} \right) \end{align}

What about

\begin{align} \left( \frac{1}{1+x^2 y^2} \right)^{(2n)} \end{align}

Answer 1

The first thing to do is make this substitution $y = ik$, we can do this freely because the derivative is with respect to x not to y, and it is only a contsant.Then after this substitution we get

$$\left( \frac{1}{1+x^2 y^2} \right)^{(2n)}=\left( \frac{1}{1-x^2 k^2} \right)^{(2n)} $$

Which, as you said, is equal to $$\frac{(2n)!k^{2n}}{2 } \left( \frac{1}{(1+xk)^{2n+1}} +\frac{1}{(1-xk)^{2n+1}} \right) $$

Finally we just substitute $\frac{y}{i}=k$ and we obtain that for the derivative of a number with the form 4k we have $$\frac{(2n)!y^{2n}}{2 } \left( \frac{1}{(1+\frac{xy}{i})^{2n+1}} +\frac{1}{(1-\frac{xy}{i})^{2n+1}} \right) $$ and for the derivative of a number with the form 4k+2 we have $$\frac{-(2n)!y^{2n}}{2 } \left( \frac{1}{(1+\frac{xy}{i})^{2n+1}} +\frac{1}{(1-\frac{xy}{i})^{2n+1}} \right) $$ then if we do not want i, we only have to add the two terms and apply the binomial theorem.