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.
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.