How can we differentiate $(x^{-1})^{({x^{-1})^{x^{-1}}}}$ wrt $x$?

Question

How can we differentiate $(x^{-1})^{({x^{-1})^{x^{-1}}}}$ with respect to $x$?

Answer 1

Well, for positive real $x,$ we can use the fact that $\alpha^\beta=\exp(\beta\ln\alpha)$ for any positive real $\alpha$ and any real $\beta,$ together with the power rule of logarithms, to see that $$\begin{align}\left(x^{-1}\right)^{\left(x^{-1}\right)^{x^{-1}}} &= \exp\left(\left(x^{-1}\right)^{x^{-1}}\ln\left(x^{-1}\right)\right)\\ &= \exp\left(-\left(x^{-1}\right)^{x^{-1}}\ln(x)\right)\\ &= \exp\left(-\exp\left(x^{-1}\ln\left(x^{-1}\right)\right)\ln(x)\right)\\ &= \exp\left(-\exp\left(-x^{-1}\ln(x)\right)\ln(x)\right).\end{align}$$

After that, it simply becomes an exercise in repeated use of the chain rule, together with the product rule.

Answer 2

$$(x^{-1})^{({x^{-1})^{x^{-1}}}}=(x^{-1})^{({x^{-1/x})}}=x^{-x^{-1/x}}=y$$ $$\ln y=-x^{-1/x}\ln x=z\ln x$$ $$y'=y(z'\ln x+\frac{z}{x})$$ where $$-x^{-1/x}=z$$ $$\ln z=\frac{1}{x}\ln x\Rightarrow z'=z\frac{1-\ln x}{x^2}$$

Answer 3

spurred on by Cameron's excellent approach, I'm wondering about:

let $z=\frac1x$, so $$f(x) = z^{z^z}$$ and $$f'(x) = \frac{d}{dz} z^{z^z} \frac{dz}{dx} = -x^{-2}\frac{d}{dz} z^{z^z} $$ as a possible alternative route?