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 $x$
$$y_{1}=x.((-1).(x^2+a^2)^{-2}.2x)+(x^2+a^2)^{-1}$$
$$y_{1}=(-1).(x^2+a^2)^{-2}.2x^2+(x^2+a^2)^{-1}$$

Hint:
Rewrite the definition of $y$ as $(x^2+a^2)y=x$, and apply Leibniz' formula for the $n$-th derivative of a product of functions: for $n>1$, we have
$$(x^2+a^2)y^{(n)}+2nx  y^{(n-1)}=0,$$
whence the recurrence relation
$$y^{(n)}=-\frac{2nx}{x^2+a^2}y^{(n-1)}.$$
