Question

$y=\frac{1}{1+x+x^2+x^3}{\left[\frac{d{y}^2}{d{y}^2}\right]}_{x=0}=?$

Answer 1

$y=\frac{1}{1+x+x^2+x^3}$

$\frac{dy}{dx}=\frac{(1+x+x^2+x^3)\left(0\right)-\left(1\right)(1+2x+3x^2)}{{(1+x+x^2+x^3)}^2}=-\frac{(1+2x+3x^2)}{{(1+x+x^2+x^3)}^2}$

$\frac{d^{2}y}{d{x}^2}=-\left[\frac{{(1+x+x^2+x^3)}^2(2+6x)-2(1+2x+3x^2)\left((1+x+x^2+x^3)\right)(1+2x+3x^2)}{{(1+x+x^2+x^3)}^4}\right]$

${\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=0}=-\left[\frac{{\left(1\right)}^2\left(2\right)-2\left(1\right)\left(1\right)\left(1\right)}{{\left(1\right)}^4}\right]=0$