Find $\frac{dy}{dx}$
$\begin{align*} y&=\frac{x^2-1}{x^4-1}\\ &=\frac{x^4-1(2x)-x^2-1(4x^3)}{(x^4-1)^2}\\ &=\frac{2x^5-2x-4x^5-4x^3}{(x^4-1)^2} \end{align*}$ but the right answer is $\frac{-2x^5+4x^3-2x}{(x^4-1)^2}$ what did I do right, I used quotient rule.
I want to use below formula, but i don't know how to
$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
many thanks in advance!