Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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…. – Américo Tavares Dec 4 '12 at 20:22
up vote 7 down vote accepted

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}

share|cite|improve this answer
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
Sorry, I can see now. – user29999 Dec 4 '12 at 19:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.