Find the $nth$ derivative of $\frac{3x-1}{x^2+x+1}.$
I tried solving this problem by simplifying the given expression $\frac{3x-1}{x^2+x+1}.$ Now, $\frac{3x-1}{x^2+x+1}=\frac{\frac 13 (x-\frac 13)}{x^2+x+1}.$ So, if we are able to find the $nth$ derivative of $\frac{x-\frac 13}{x^2+x+1}=\frac{x-\frac 13}{(x+\frac 12)^2+\frac 34},$ then we are done.But I am not finding any productive way of solving or proceeding it further as it doesn't give us something helpful such that it reduces to one of the standard form whose nth derivatives are known. I am not quite getting it.
Best Answer
Find the roots of $x^2+x+1$ and apply partial fractions. We will call the roots of the denominator $a,b$ $$\frac{3x-1}{x^2+x+1}=\frac{3x-1}{(x-a)(x-b)}=\frac{3a-1}{a-b}\frac{1}{x-a}-\frac{3b-1}{a-b}\frac{1}{x-b}$$ $$\left(\frac{d}{dx}\right)^n\frac{1}{x-a}=\frac{(-1)^n\space n!}{(x-a)^{n+1}}$$ $$\left(\frac{d}{dx}\right)^n\frac{3x-1}{x^2+x+1}=\frac{3a-1}{a-b}\frac{(-1)^n\space n!}{(x-a)^{n+1}}-\frac{3b-1}{a-b}\frac{(-1)^n\space n!}{(x-b)^{n+1}}$$$$=\frac{(-1)^nn!}{a-b}\left(\frac{3a-1}{(x-a)^{n+1}}-\frac{3b-1}{(x-b)^{n+1}}\right)$$ $$a,b=\frac{-1±i\sqrt3}{2}$$ We can easily induce a general strategy by following the same method. We start by factoring the denominator and applying partial fractions, allowing us to use the formula from earlier $$\frac{mx+b}{(x-r)(x-s)}=\frac{mr+b}{r-s}\frac{1}{x-r}-\frac{ms+b}{r-s}\frac{1}{x-s}$$ $$\left(\frac{d}{dx}\right)^n\frac{mx+b}{(x-r)(x-s)}=\frac{mr+b}{r-s}\frac{(-1)^n\space n!}{(x-r)^{n+1}}-\frac{ms+b}{r-s}\frac{(-1)^n\space n!}{(x-s)^{n+1}}$$