I can't find the derivative of:
f(x) = (2x^2) / (x^2-1)
Having a hard time with this one.
First way: quotient rule: $\displaystyle f'(x)=\frac{4x(x^2-1)-2x(2x^2)}{(x^2-1)^2}=\frac{-4x}{(x^2-1)^2}$
Second way: first write $\displaystyle f(x)=2+\frac{2}{x^2-1}$ and now we only need the derivative of $\displaystyle \frac{1}{x^2}$ :
$\displaystyle f'(x)=\frac{-4x}{(x^2-1)^2}$
Tonio