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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!

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    The lack of parentheses is messing you up.2012-04-22

1 Answers 1

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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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    my bad, thx alot! @anon2012-04-22