<s>[INST] Find the $n^{th}$ derivative of $y=\frac {x}{x^2+a^2}$ Find the $n^{th}$ derivative of $y=\dfrac {x}{x^2+a^2}$.
My Attempt:
$$y=\dfrac {x}{x^2+a^2}$$
$$y=x.(x^2+a^2)^{-1}$$
Differentiating both sides with respect to $x$
$$y_{1}=x.((-1).(x^2+a^2)^{-2}.2x)+(x^2+a^2)^{-1}$$
$$y_{1}=(-1).(x^2+a^2)^{-2}.2x^2+(x^2+a^2)^{-1}$$
 [/INST] $$y=\frac{x}{x^2+a^2}=\frac{1}{2} \left( \frac{1}{x-ia}+\frac{1}{x+ia} \right)$$
The nth derivative of $(x+b)^{-1}$ is given as
$$D^n (x+b)^{-1}=(-1)^n n!(x+b)^{-n-1}$$
Using this
$$D^n y= (-1)^n n!~ \Re (x+ia)^{-n-1}=(-1)^n n! (x^2+a^2)^{(-n-1)/2} \cos[(n+1) \tan^{-1}(a/x)].$$
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