2

I am trying to solve:

$x^{\sqrt{y}}=y^{\sqrt{x}}$

Here's my solution. Please correct me if there is an error and kindly explain why.

$\sqrt y\ln{x}=\sqrt x\ln{y}$

$\left[y^\frac{1}{2}\ln{x}\right]^\prime=\left[x^\frac{1}{2}\ln{y}\right]^\prime$

$\frac{1}{2}y^{-\frac{1}{2}}\ y^\prime\ln{x}+y^\frac{1}{2}\left(\frac{1}{x}\right)=\frac{1}{2}\ x^{-\frac{1}{2}}\ \ln{y}+x^\frac{1}{2}\left(\frac{1}{y}\right)y^\prime$

$\frac{\ln{x}}{2\sqrt y}\ y^\prime+\frac{\sqrt y}{x}=\frac{\ln{y}}{2\sqrt x}\ +\frac{\sqrt x}{y}y^\prime$

$\ \left(\frac{\ln{x}}{2\sqrt y}-\frac{\sqrt x}{y}\right)y^\prime=\frac{\ln{y}}{2\sqrt x}-\frac{\sqrt y}{x}\ $

$\left(\frac{y\ln{x}-2\sqrt{xy}}{2y\sqrt y}\right)y^\prime=\frac{x\ln{y}-2\sqrt{xy}}{2x\sqrt x}$

$y^\prime=\frac{y\sqrt y\left(x\ln{y}-2\sqrt{xy}\right)}{x\sqrt x\left(y\ln{x}-2\sqrt{xy}\right)}$

Thanks in advance.

PRD
  • 611
  • Hi :) How about $y(x) =x$ for all $x>0$, then $y'(x) =1$. – Jochen Nov 14 '23 at 06:05
  • 1
    @Jochen If you mean your comment as support for OP result, then OK. It's a special case of the result in the question. [A simpler response might be that the OP calculation seems to be correct.] – coffeemath Nov 14 '23 at 06:10
  • @Jochen. What you wrote is not true (think about Lambert function). Try with $x=2$ and $y=80$; the lhs is $492.600$ and the rhs is $491.333$ – Claude Leibovici Nov 14 '23 at 06:43
  • 1
    the curve is not differentiable at $(x,y)=(e^2,e^2)$, other than that Your solution looks correct – Dheeraj Gujrathi Nov 14 '23 at 10:50
  • Oh ok yes, I am sorry – Jochen Nov 14 '23 at 17:32
  • @DheerajGujrathi - the particular equation produced fails at $x = y = e^2$. However, that does not mean the curve $y(x)$ is not differentiable there. Indeed, the simplest solution, $y = x$, obviously is. – Paul Sinclair Nov 14 '23 at 21:39
  • @PaulSinclair,y'=1 is just one part of solution,every point $(x,y)$ on the curve(Which actually looks like bow and arrow) except $(e^2,e^2)$ has a local Derivative which can be written in form of the expression obtained At end By OP,It just happens to be the case that the x=y line is part of the curve ,Hence derivative is 1 at all such points,But general solution is still that was obtained at end – Dheeraj Gujrathi Nov 14 '23 at 21:58
  • @DheerajGujrathi - I guess you are talking about the graph of all points $(x, y)$ satisfying the equation. But the concept of $y'$ requires thinking $y$ as a function of $x$ instead, for which there are two differentiable solutions of maximal domain. $y = x$ on $(0,\infty)$ is one, and the other on $(1,\infty)$ with a more complicated form. These two curves cross at $(e^2, e^2)$ (where your combined concept fails to be a curve), which is why the implicit derivative expression which applies to both is necessarily not defined there. However each curve itself is differentiable at that point. – Paul Sinclair Nov 15 '23 at 05:18
  • @PaulSinclair, "However each curve itself is differentiable at that point",even if These two are two different curves, They are only different parts of a non functional relation satisfying $x^{\sqrt y}=y^{\sqrt x}$, Being more rigorous, Defining derivatives for a curve following certain relation is itself tricky, consider circle, It has well defined derivative at each point on the curve "locally", The closest rigorous definition I can get on MSE is as below – Dheeraj Gujrathi Nov 15 '23 at 06:09
  • @PaulSinclair,>:Consider a relation $(x,y)=R\subseteq\Bbb{R}\times\Bbb{R}$ and any point $(x_0,y_0)\in\Bbb{R}\times\Bbb{R}$, we can give a tentative definition of the derivative as follows,

    $$\left.\frac{dy}{dx}\right|{(x_0,y_0)}=\lim{\substack{(x_1,y_1)\rightarrow (x_0,y_0)}\(x_1,y_1)\in R\setminus {x_0}\times\Bbb{R}}\frac{y_1-y_0}{x_1-x_0}$$

     – Dheeraj Gujrathi Nov 15 '23 at 06:09

0 Answers0