$\displaystyle y=(\frac{x^2+1}{x^2-1})^3$
I used quotient rule and chain rule.
The solution keep getting is wrong. I think my algebra is off.: $\displaystyle y'= \frac{12x(x^2+1)^2}{(x^2-1)^4}$
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
$\displaystyle y'=3(\frac{x^2+1}{x^2-1})^2$
$\displaystyle \displaystyle y=\left(\frac{x^2+1}{x^2-1}\right)^3$
$\displaystyle \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{(x^2-1)(2x) - (x^2+1)(2x)}{(x^2-1)^2}$
$\displaystyle \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{2x[x^2-1 - (x^2+1)]}{(x^2-1)^2}$
$\displaystyle \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{-4x}{(x^2-1)^2}$