Where is the mistake in the calculation of $y'$ if $y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4}$?

Plase take a look here.

If $y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4}$

\begin{eqnarray} y'&=& \dfrac{1}{4} \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \left \{ \dfrac{2x(x^2-1) - 2x(x^2+1) }{(x^2-1)^2} \right \}\\ &=& \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}. \end{eqnarray} By the other hand, we have $$\log y = \dfrac{1}{4} \left \{ \log (x^2+1) - \log (x^2-1) \right \}$$ Then, \begin{eqnarray} \dfrac{dy}{dx} &=& y \dfrac{1}{4} \left \{ \dfrac{2x}{(x^2+1)} -\dfrac{ 2x}{(x^2-1)} \right \} \\ &=& \dfrac{1}{4} \dfrac{x^2+1}{x^2-1} \cdot 2x \dfrac{(x^2-1) - (x^2+1)}{(x^2+1)(x^2-1)} \\ &=& \dfrac{x^2+1}{x^2-1} \dfrac{-x}{(x^2+1)(x^2-1)} \\ &=& \dfrac{-x}{(x^2-1)^2}. \end{eqnarray} But this implies, $$\dfrac{-x}{(x^2-1)^2} = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}.$$ Where is the mistake?

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It's recommendable that you use LaTeX in the exponents. Instead of $x²$, use $x^2$ x^2. Also, there's a typo in title "calculation". Better $y'$ than $y´$. – Américo Tavares Dec 4 '12 at 19:49
@AméricoTavares : I was about to post the same comment about squares. We had that same discussion several years ago on Wikipedia, about the style manual for typesetting in math articles. – Michael Hardy Dec 4 '12 at 20:20
@MichaelHardy I saw your post on meta meta.math.stackexchange.com/questions/6717/…. – Américo Tavares Dec 4 '12 at 20:22

I believe you forgot a power 1/4 when substituting for $y$ (in the calculation using logarithms).
I think no, look $y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} \Rightarrow \log y = \dfrac{1}{4} \log \left \{ \dfrac{x^2+1}{x^2-1} \right \} = \dfrac{1}{4} \left \{ \log (x²+1) - \log (x²-1) \right \}$ – user29999 Dec 4 '12 at 19:53
When substituting for $y$. This is from the first to the second line of the calculation starting with $\frac{dy}{dx}$. – Daan Michiels Dec 4 '12 at 19:54