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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 \begin{equation} \log y = \dfrac{1}{4} \left \{ \log (x^2+1) - \log (x^2-1) \right \} \end{equation} 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, \begin{equation} \dfrac{-x}{(x^2-1)^2} = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}. \end{equation} Where is the mistake?

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    $\begingroup$ 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 $y´$. $\endgroup$ Commented Dec 4, 2012 at 19:49
  • $\begingroup$ @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. $\endgroup$ Commented Dec 4, 2012 at 20:20
  • $\begingroup$ @MichaelHardy I saw your post on meta meta.math.stackexchange.com/questions/6717/…. $\endgroup$ Commented Dec 4, 2012 at 20:22

1 Answer 1

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I believe you forgot a power 1/4 when substituting for $y$ (in the calculation using logarithms).

Edited to explain further: In your calculation, you write \begin{align} \frac{dy}{dx} &= y\frac14 \left\{ \frac{2x}{(x^2+1)} - \frac{2x}{(x^2-1)} \right\} \\ &= \frac14 \frac{x^2+1}{x^2-1} \cdot 2x\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)}. \end{align} However, this should be \begin{align} \frac{dy}{dx} &= y\frac14 \left\{ \frac{2x}{(x^2+1)} - \frac{2x}{(x^2-1)} \right\} \\ &= \frac14 \color{red}{\left(\color{black}{\frac{x^2+1}{x^2-1}}\right)^{\frac14}} \cdot 2x\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)}. \end{align}

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  • $\begingroup$ 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 \} $ $\endgroup$
    – user29999
    Commented Dec 4, 2012 at 19:53
  • $\begingroup$ When substituting for $y$. This is from the first to the second line of the calculation starting with $\frac{dy}{dx}$. $\endgroup$ Commented Dec 4, 2012 at 19:54
  • $\begingroup$ Sorry, I can see now. $\endgroup$
    – user29999
    Commented Dec 4, 2012 at 19:59

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