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In my post A very interesting question: intersection point of $x^y=y^x$ I demonstrated that for the graph $x^y=y^x$, $$\frac{dy}{dx}=\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}$$ HOWEVER

The following point has been brought to my attention by a friend: if we were to instead to calculate the derivative like this: $$x^y=y^x\implies\frac{y}{\ln y}=\frac{x}{\ln x}\implies\frac{dy}{dx}\left(\frac{\ln y-1}{\ln^2y}\right)=\frac{\ln x-1}{\ln^2x}$$ $$\implies\frac{dy}{dx}=\frac{\ln^2y~(\ln x-1)}{\ln^2x~(\ln y-1)}$$ So what we find is that if $x^y=y^x$ then $$\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}=\frac{\ln^2y~(\ln x-1)}{\ln^2x~(\ln y-1)}$$

How do I prove this algebraically (ie NOT using calculus)? The only possible progress I have made is to observe that $$\frac{\ln^2y}{\ln^2x}=\frac{y^2}{x^2}$$ on noting that $$\frac{\ln^2y}{\ln^2x}=(\log_x y)^2=\frac{y^2}{x^2}$$ Thank you for your help. It is very much appreciated.

A-Level Student
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1 Answers1

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For $x^y=y^x$, we have $y \ln x = x \ln y$ and $\dfrac {\ln^2 y}{\ln^2 x} = \dfrac {y^2}{x^2}$, as you have observed. Now:

\begin{align}&\quad\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}\\&= \frac{y^x \ln y -\frac yxx^y}{x^y \ln x -\frac xy y^x}\\&= \frac{y^x (\ln y -\frac yx)}{x^y (\ln x -\frac xy)}&\quad x^y=y^x\\&= \frac{\ln y -\frac yx}{\ln x -\frac xy}\\&= \frac{xy\ln y -y^2}{xy\ln x -x^2}\\&= \frac{y^2\ln x -y^2}{x^2\ln y -x^2}&\quad y \ln x = x \ln y\\&= \frac{y^2(\ln x -1)}{x^2(\ln y -1)}\\&=\frac{\ln^2y~(\ln x-1)}{\ln^2x~(\ln y-1)}&\quad \frac {\ln^2 y}{\ln^2 x} = \frac {y^2}{x^2}\end{align}

player3236
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