Problem:$$\frac{\mathrm d^{99}}{\mathrm dx^{99}}Arsh(x)$$
Without: Taylor  

My attempt:
$$y=Arsh(x)$$
$$y'=\frac{1}{1+x^2}=(1+x^2)^{-\frac{1}{2}}=t$$
$$y''=\frac{-x}{(1+x^2)^{-\frac{3}{2}}}=\frac{-x\*t}{(1+x^2)}$$
$$-x\*y'=y''\*(1+x^2)$$ 
And if I use general Leibniz rule on that with n=98 but I can't get just 100st
derivative of y(x).