A follow-up question to the previous question.

Is there a general formula of the first order derivative of

$$ \ln(x)↑↑n$$


Where the $n$ is a constant independent of $x$.


closed as off-topic by user223391, Leucippus, JonMark Perry, Rohan, Claude Leibovici Dec 31 '17 at 10:19

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  • 1
    $\begingroup$ There is no question here $\endgroup$ – user223391 Dec 30 '17 at 16:58
  • $\begingroup$ Can you please write a more understandable question? $\endgroup$ – user507623 Dec 30 '17 at 16:58
  • $\begingroup$ I think he means: $y = \ln (x)$ and we have to find $\frac d{dx} y^y^y^y^y^...y$. $n$ times. $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 at 17:01
  • $\begingroup$ @MohammadZuhairKhan yes that is correct $\endgroup$ – Fred Wieser Dec 30 '17 at 17:02
  • $\begingroup$ sorry for the confusion $\endgroup$ – Fred Wieser Dec 30 '17 at 17:02

Consider the function $f(x,y) = y^x$. If one considers $x, y$ to be functions of some variable $t$, then the multivariate chain rules says $$\begin{align}\frac{df}{dt} &= \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}\\&=(y^x\ln y)x' + (xy^{x-1})y'\\&=f(x,y)\left(x'\ln y + y'\frac xy\right)\end{align}$$

Now consider $g_n(x) = x↑↑n$ and $h_n(x) = \ln g_n(x)$. Clearly $g_n(x) = f(x,g_{n-1}(x))$, so $$\frac{dg_n}{dx} = g_n(x)\left(1\cdot\ln g_{n-1}(x) + \frac{dg_{n-1}}{dx}\frac{x}{g_{n-1}(x)}\right)$$ I.e., $$\frac{d(\ln g_n)}{dx} = \ln g_{n-1}(x) + x\frac{d(\ln g_{n-1})}{dx}\\h_n' = h_{n-1} + xh_{n-1}'$$

This recursion formula can be used to find the logarithmic derivative of $g_n(x) = x↑↑n$ for any $n$. And the normal derivative is obtained by multiplying by $x↑↑n$.

Finally, the derivative you are asking for is of $g_n(\ln x)$, which can be obtained by the chain rule.


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