Differentiation of an implicit function

Question

$$ x = y^x $$

been working through some implicit functions and this has me a bit, I don't think my answer is correct

I have treated $y$ as $y(x)^x$ and differentiated with respect to x

Answer 1

Let $$ F(x,y) = y^x - x = 0. $$ Then $$ \frac{dy}{dx} = - \frac{\partial F/\partial x}{\partial F/\partial y} = -\frac{y^x \ln y - 1}{xy^{x-1}} $$ We can simplify this: $y^x = x$, and $$ \frac{dy}{dx} = -\frac{x\ln y - 1}{x^2 / y} = \frac{y}{x^2} - \frac{y\ln y}{x} $$ We can take $\log$: $$ x\ln y = \ln x\Longrightarrow \ln y = \frac{\ln x}{x}\tag1 $$ So, $$ \frac{dy}{dx} = \frac{y}{x^2}(1-\ln x) $$

Or, we may follow @Mattos and differentiate $(1)$ directly.

There is another way. We know that $$ y = x^{1/x} = e^{\ln x/x}. $$ So, $$ y ' = x^{1/x} \Big(\frac{\ln x}{x}\Big)' = x^{1/x}\frac{1-\ln x}{x^2} = \frac{y}{x^2} (1-\ln x). $$