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Ok, so this question is a bit odd: it is asking for the slope of the tangent line to the function

$$x^y = y^x$$

at the points $(c,c)$. Yeah, the variable is $c$, I'm somewhat confused. I plugged in to the implicitly derived function

$$\frac{dy}{dx}=\frac{y(y-x\ln(y))}{x(x-y\ln(x)}$$

and got this

$$\frac{dy}{dx}=\frac{c(c-c\ln(c))}{c(c-c\ln(c)}$$

I'm not sure if I'm overthinking it and the answer is simply 1.

PM 2Ring
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Carlos V
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5 Answers5

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The equation $x^y = y^x$ is a bit funny, as you can see from the graph, which looks roughly like a hyperbola, with the line $y=x$ added to it. The line crosses the "hyperbola" where $y=x=e$, so the derivative at that point is not well-defined.

graph of x^y = y^x]

(Graph courtesy of Wolfram Alpha)

Clearly, for any other point $(c,c)$, i.e., points where $c \ne e$, the slope of the tangent to $x^y = y^x$ at $(c,c)$ is simply the slope of the line $y=x$, which is 1.

I suppose I ought to mention that if we remove the $y=x$ part and approach $(e, e)$ along the "hyperbolic" branch of $x^y = y^x$ the derivative approaches -1.

FWIW, although $x^y = y^x$ can't be expressed explicitly using elementary functions, it can be using the Lambert W function: $$y=\exp \left( -W \left( \frac{-\ln x}{x} \right) \right)$$

Also, it's easy to generate solution pairs $(x,y)$ of $x^y = y^x$ by this simple parametrization:

Let $y = ux$

$$\begin{align} x^y & = y^x\\ x^{ux} & = y^x\\ x^{u} & = y = ux\\ x^{u-1} & = u\\ x & = u^\frac{1}{u-1}\\ y & = u^\frac{u}{u-1}\\ \end{align}$$

PM 2Ring
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1

You need a lot more context here, but you are correct in that if you plug $(x,y) = (c,c)$ into that equation you get $\frac{dy}{dx} = 1$, or equivalently $y=x$

Brevan Ellefsen
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  • yeah i just wasnt sure if that would simply be the answer or not and regards to the context what am i missing? I can vouch for the derivative to be right. – Carlos V Sep 20 '15 at 06:18
  • @CarlosV Since you just posted about differentiating $x^y = y^x$ I was under the assumption that this was your context, although this answer is still conditional still upon the domain of $c$... I assumed both are $ > 0$, as this answer is wildly different otherwise – Brevan Ellefsen Sep 20 '15 at 06:23
  • I have another question hoping you can answer how would i determine at what points would make the tangent line not exist? – Carlos V Sep 20 '15 at 06:25
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    @CarlosV to start, $\log(0)$ is undefined, so if $c=0$ the tangent line is undefined. If you are ignoring complex numbers then the tangent line doesn't necessarily exist for $c<0$ either, although this is heavily conditional. – Brevan Ellefsen Sep 20 '15 at 06:26
  • Would that still qualify for the derivative? $\frac{dy}{dx}=\frac{y(y-x\ln(y))}{x(x-y\ln(x))}$ – Carlos V Sep 20 '15 at 06:29
  • I'm not sure what you mean by "qualify" here. I specified that $c=0$ in my comment, and the derivative does not have a $c$ in it (as $x$ and $y$ can differ). Do you mind elaborating on what you mean? It is true that the derivative is undefined if either $x=0$ or $y=0$, although the limit of the derivative as $x \to 0$ exists (and analogously $y \to 0$), as (as $\lim {x \to 0} [\log(x)]= -\infty$). – Brevan Ellefsen Sep 20 '15 at 06:34
  • Let us continue this discussion in chat. – Carlos V Sep 20 '15 at 06:36
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    @CarlosV: That is a good derivative. But only at those points where the implicit function theorem can be applied. For example that formula gives you the slope of the tangent to the curve at $(x,y)=(2,4)$. But you get $0/0$ at $(x,y)=(e,e)$. The implicit function theorem does not guarantee you a tangent at that point, and the geometric reason for that is apparent from PM 2Ring's answer below - two branches meet there. – Jyrki Lahtonen Sep 20 '15 at 09:35
  • Slope $dy/dx$ at (0,0) and (e,e) appears be zero, in the immediate neighborhood of (e,e). Verified by plot below also. – Narasimham Sep 20 '15 at 18:34
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Notice, logarithms is defined for positive real numbers only hence you need to give a condition of arbitrary constant $\color{red}{c>0}$ otherwise log will be undefined.

Moreover the slope will be undefined in form of $\frac{0}{0}$ at $c=e$ so we should have another condition $\color{red}{c\neq e}$

Now, we have $$\frac{dy}{dx}=\frac{y(y-x\ln(y))}{x(x-y\ln(x))}$$

Now, the slope of the tangent at the given point $(c, c)$ is $$\left(\frac{dy}{dx}\right)_{(x=c, \ y=c)}=\frac{c(c-c\ln(c))}{c(c-c\ln(c))}=1$$

so your answer is correct (given additional conditions $\color{red}{c>0, \ c\neq e}$).

Harish Chandra Rajpoot
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  • oops yeah i forgot about that. – Carlos V Sep 20 '15 at 06:22
  • This is not true. Logarithms can be defined for negative numbers easily, and even complex numbers. You just either get 1 or infinite solutions, unless you make a branch cut. – Brevan Ellefsen Sep 20 '15 at 06:25
  • Alright, if you are right then $\ln(-2)=$? & $\ln(2+i3)=$? – Harish Chandra Rajpoot Sep 20 '15 at 06:28
  • @HarishChandraRajpoot Assuming the standard branch cut, $\log(-2) = \log(2) + \pi i$, and $\log(2+3i)$ is a transcendental number (i.e., it can't be simplified further, but is $\approx$ $1.28i$) – Brevan Ellefsen Sep 20 '15 at 06:31
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    @HarishChandraRajpoot Actually, you can write $\log(2+3i)$ as $\frac{\log(13)}{2}+i \tan^{-1}(\frac{3}{2})$ if you want, but this isn't really any simpler – Brevan Ellefsen Sep 20 '15 at 06:38
  • @BrevanEllefsen: Technically you're right, but surely in the context of this problem we are restricted to real $x,y$ (necessarily positive for the power functions to be well defined). Thus I fail to see the point of bringing up complex logarithms here. – Jyrki Lahtonen Sep 20 '15 at 09:31
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I think that the main issue here is whether the fraction can always be simplified.

Clearly, for $c\neq e, 0$,

$$\frac{dy}{dx}=1$$.

So for continuous function $f(x)$, $\frac{dy}{dx}$ approaches to $1$ from both side at $x=e$.

Similar for $x=0$.

Mythomorphic
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  • You bring up an interesting point here that $c$ is undefined for $c=e$. Good catch! – Brevan Ellefsen Sep 20 '15 at 06:28
  • We may perhaps need to find a limiting function $ \frac{x^y-y^x}{x-y} $ as $ x \rightarrow y$. I asked a doubt here related to this couple of days ago. http://math.stackexchange.com/questions/1439390/limit-of-function-formed-by-taking-difference-after-swapping-a-variable-consta – Narasimham Sep 20 '15 at 12:29
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HINT:

If we attempt to find solution for $ e^x = x^e $ we have a second real root in addition to the obvious $ x= e$.

Similarly $ x^y = y^x $ contains another real function other than the obvious $ x= y, $ that should be isolated.

$ dy/dx for x^y-y^x $

Narasimham
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