How do I differentiate $$ x^{x^{x^{…}}}$$ with respect to $x$? (Note that $x$ is raised infinitely many times.)
My attempt: Let $y = x^{x^{x^{…}}}$. Taking logarithm of both sides we get $\ln y = y \ln x$ and let $f = y \ln x – \ln y$. Now $$\frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = \frac{y^2}{x(1 – y \ln x)}$$
Is this approach correct? If not how do I proceed ?
Help would be appreciated. Thanks.
Best Answer
The way I would do this: $$y=x^{x^{x^{\cdots}}}=x^y=\exp(y\ln x)$$ where I've used the notation $\exp(...)$ instead of $e^{\cdots}$ to make the working neater. Now, differentiating implicitly with respect to $x$, and using the fact that $\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$gives $$\begin{align}\frac{dy}{dx}&=\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)\exp(y\ln x)\\ &=\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)x^y\end{align}$$ Rearrange to isolate $\frac{dy}{dx}$: $$\frac{dy}{dx}\left(1-x^y\ln x\right)=\frac{yx^y}{x}\implies\frac{dy}{dx}=\frac{yx^y}{x(1-x^y\ln x)}=\frac{y^2}{x(1-y\ln x)}$$
which is the same result that you got; your way seems fine :)
I hope that was useful. If you have any questions please don't hesitate to ask :)