Question

If $y=\frac{x}{x^2-1}$ then $-y_5\left(0\right)$ is equal to

Answer 1

$y=\frac{x}{x^2-1}y=\frac{x}{(x+1)(x-1)}2y=\frac{1}{x+1}+\frac{1}{x-1}{2y}_1=-\frac{1}{(x+1{)}^2}-\frac{1}{(x-1{)}^2}{2y}_2=\frac{2}{(x+1{)}^3}+\frac{2}{(x-1{)}^3}{2y}_3=-\frac{6}{(x+1{)}^4}-\frac{6}{(x-1{)}^4}{2y}_4=\frac{24}{(x+1{)}^5}+\frac{24}{(x-1{)}^5}{2y}_5=-\frac{120}{(x+1{)}^6}-\frac{120}{(x-1{)}^6}{2y}_5\left(0\right)=-\frac{120}{(0+1{)}^6}-\frac{120}{(0-1{)}^6}\Rightarrow {-y}_5\left(0\right)=120$