Find slope at tangent line.

Question

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.

Answer 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 $y=x$

Answer 2

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 $\color{red}{c>0, \ c\neq e}$).

Answer 3

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 $x=0$.

Answer 4

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 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.

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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 $y = ux$

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

Answer 5

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.