# 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?

• 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 $y´$. Commented Dec 4, 2012 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. Commented Dec 4, 2012 at 20:20
• @MichaelHardy I saw your post on meta meta.math.stackexchange.com/questions/6717/…. Commented Dec 4, 2012 at 20:22

I believe you forgot a power 1/4 when substituting for $y$ (in the calculation using logarithms).
• I think no, look $y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} \Rightarrow \log y = \dfrac{1}{4} \log \left \{ \dfrac{x^2+1}{x^2-1} \right \} = \dfrac{1}{4} \left \{ \log (x²+1) - \log (x²-1) \right \}$ Commented Dec 4, 2012 at 19:53
• When substituting for $y$. This is from the first to the second line of the calculation starting with $\frac{dy}{dx}$. Commented Dec 4, 2012 at 19:54