# 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
$$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$

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If $y=\frac{x^2-1}{x^4-1}$, then $y$ cannot simultaneously equal $\frac{x^4-1(2x)-x^2-1(4x^3)}{(x^4-1)^2}$. Did you mean for some of your $y$s to be $\frac{dy}{dx}$? – Henning Makholm Apr 22 '12 at 10:06
The lack of parentheses is messing you up. – Arturo Magidin Apr 22 '12 at 21:48

The quotient rule is

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f\,'(x)g(x)-f(x)g'(x)}{g(x)^2}.$$

$$f(x) = x^2-1; \quad g(x)=x^4-1$$

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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yea. i did it upto this step but, the answer is not right – Sb Sangpi Apr 22 '12 at 10:11
@Sb: You were not careful with your parentheses! (See the edit.) – anon Apr 22 '12 at 10:17
my bad, thx alot! @anon – Sb Sangpi Apr 22 '12 at 10:34