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I would like to work out the sign of the first order derivative of $y= (\frac{x}{x-1})^x - \frac{x}{x-1}$ with respect to x.

I get this: $dy/dx = (\frac{x}{x-1})^x \left(-\frac{1}{x-1} + ln\frac{x}{x-1} + (\frac{1}{x-1})^2 (\frac{x-1}{x})^x \right)$

when I plot this function, for $x>1$, I see that as $x$ increases, $y$ decreases.

I understand that $ln\frac{x}{x-1} \leq \frac{1}{x-1}$, but we have this extra term $(\frac{1}{x-1})^2 (\frac{x-1}{x})^x$, how do I prove that $dy/dx < 0$?

mijia
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1 Answers1

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This is not a full answer.

Writing the derivative in a different way, we have $$y'=\frac{1+(x-1) \left(\frac{x}{x-1}\right)^x \left((x-1) \log \left(\frac{x}{x-1}\right)-1\right)}{(x-1)^2}$$ So, the problem is "just" the sign of the numerator.

Expanding as a Taylor series around $x=1$, we have $$y'=-(x-1)+\frac{1}{2} (x-1)^2 \log ^2(x-1)+O\left((x-1)^3\right)$$ WHich is negative at least for small values of $x$.

Similarly, for large values of $x$, $$y'=\left(1-\frac{e}{2}\right)+\frac{e}{12 x}+O\left(\frac{1}{x^2}\right)$$ which is also negative.

Now, what happens between these two points ?

Claude Leibovici
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