<s>[INST] Find $\frac{dy}{dx}$ when $y=\frac{x^2-1}{x^4-1}$ 
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!
 [/INST] The quotient rule is
$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f\,'(x)g(x)-f(x)g'(x)}{g(x)^2}.$$
In your case,
$$f(x) = x^2-1; \quad g(x)=x^4-1$$
and the derivatives are
$$f\,'(x)=2x; \quad g'(x)=4x^3.$$
Plugging everything in, with $y=f(x)/g(x),$ we have
$$\frac{dy}{dx}=\frac{(2x)(x^4-1)-(x^2-1)(4x^3)}{(x^4-1)^2}=\cdots $$

To simplify the numerator:
$$(2x)(x^4-1)-(x^2-1)(4x^3)=(2x^5-2x)-(4x^5-4x^3)$$
$$=-2x^5+4x^3-2x.$$
Did you follow that? The issue you have is you don't have any parentheses around $x^2-1$, so a sign ended up remaining when it was supposed to turn positive (due to $-1\times-1=+1$).
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