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did I make a mistake? when trying to find the derivative of $f(x)=x^{x^{x^{x^{x^{\dots}}}}}$

the first thing I did was to change the $f(x)$ to this $$y=x^y$$

I took the derivative of both $y$ and $x$ to get $$dy=yx^{y-1}dx+\ln(x)x^ydy$$

I moved all the $dy$'s onto one side $$(1-\ln(x)x^y)dy=yx^{y-1}dx$$

$$\frac{(1-\ln(x)x^y)dy}{dx}=yx^{y-1}$$

$$\frac{dy}{dx}=\frac{yx^{y-1}}{(1-\ln(x)x^y)}$$

So I got $$\frac{d}{dx} x^{x^{x^{x^{x^{\dots}}}}}=\frac{x^{x^{x^{x^{x^{\dots}}}}}x^{x^{x^{x^{x^{x^{\dots}}}}}-1}}{(1-\ln(x)x^{x^{x^{x^{x^{x^{\dots}}}}}})}$$

$$\frac{d}{dx} x^{x^{x^{x^{x^{\dots}}}}}=\frac{x^{2x^{x^{x^{x^{x^{\dots}}}}}-1}}{(1-\ln(x)x^{x^{x^{x^{x^{x^{\dots}}}}}})}$$

I feel like I did something wrong when solving it.

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    here is a relevant question. This just considers $x^x$ but it handles it in a way analogous to yours. – lulu Apr 12 '21 at 02:11

1 Answers1

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An alternative approach.

After $y = x^y$

Proceed as follows:

$y = e^{y \ln x} $

Differentiate implicitly with respect to $x$:

$y' = e^{y \ln x} \cdot (y' \ln x + \frac yx) = x^y \cdot (y' \ln x + \frac yx) $

Now group the $y' = \frac{dy} {dx} $ terms together.

Deepak
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  • @lulu Thank you. I am out now but I'll take a closer look when I'm home. Meanwhile, I'll edit the post to retract my statement about the error in the OP's work and state this is an alternative approach. I appreciate your picking up on this. – Deepak Apr 12 '21 at 02:47
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    I'll delete my prior comment now. As I mentioned to the OP, this point came up recently in connection with the derivative of $x^x$. There, interestingly, the OP's method was first cited as an example of a false method used to obtain a true result. – lulu Apr 12 '21 at 11:02