$\displaystyle y=(\frac{x^2+1}{x^2-1})^3$

I used quotient rule and chain rule.

The solution keep getting is wrong. I think my algebra is off.: $\displaystyle y'= \frac{12x(x^2+1)^2}{(x^2-1)^4}$

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- Jul 28th 2010, 04:24 PMsuperduper1find the derivative
$\displaystyle y=(\frac{x^2+1}{x^2-1})^3$

I used quotient rule and chain rule.

The solution keep getting is wrong. I think my algebra is off.: $\displaystyle y'= \frac{12x(x^2+1)^2}{(x^2-1)^4}$ - Jul 28th 2010, 04:29 PMAlso sprach Zarathustra
y'=3{(x^2+1)/(x^2-1)}^2 * {2x(x^2-1)-2x(x^2+1)}/{x^2-1}^2=...

}

=3{(x^2+1)/(x^2-1)}^2 * {2x{x^2-1-x^2-1}/{(x^2-1)^2}=

=3{(x^2+1)/(x^2-1)}^2 * {2x{-2}/{(x^2-1)^2}=

=3{(x^2+1)/(x^2-1)}^2 * {-4x/{(x^2-1)^2}=

=-12x/(x^2-1)}^4 - Jul 28th 2010, 04:30 PMskeeter
- Jul 28th 2010, 04:31 PMpickslides
I get $\displaystyle y = \frac{6x(x^2-1)^3(x^2+1)^2-6x(x^2-1)^2(x^2-1)^3}{(x^2-1)^6}$

take out common factors to do some additional cancellations. - Jul 28th 2010, 04:42 PMsuperduper1
I think the mistake is somewhere after here:

$\displaystyle y'=3(\frac{x^2+1}{x^2-1})^2 * \frac{2x[x^2+1-x^2+1]}{(x^2-1)^2}$

$\displaystyle y'=3(\frac{x^2+1}{x^2-1})^2 * \frac{2x(2)}{(x^2+1)^2}$

What is this when simplified? I think this is throwing me off. The 3 in front and the ^2 in the back

$\displaystyle y'=3(\frac{x^2+1}{x^2-1})^2$ - Jul 28th 2010, 05:51 PMskeeter
$\displaystyle \displaystyle y=\left(\frac{x^2+1}{x^2-1}\right)^3$

$\displaystyle \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{(x^2-1)(2x) - (x^2+1)(2x)}{(x^2-1)^2}$

$\displaystyle \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{2x[x^2-1 - (x^2+1)]}{(x^2-1)^2}$

$\displaystyle \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{-4x}{(x^2-1)^2}$