Looking at a research paper that is looking at tetrations of ln(x) and their first order derivatives. I just want someone to provide a method of how to find the derivative of $$ \ln(x)^{(\ln(x)^{\ln(x)})} $$

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  • Welcome to MSE! Try using mathjax to make your question more presentable. – Mohammad Zuhair Khan Dec 30 '17 at 16:33
  • is this $$\left((\ln(x))\right)^{\ln(x)^{\ln(x)}}$$? – Dr. Sonnhard Graubner Dec 30 '17 at 16:34
  • 1
    @Dr.SonnhardGraubner he did specify tetrations so it is clear he is not speaking of the expression you mentioned. – Isaac Browne Dec 30 '17 at 16:35
  • @Dr.SonnhardGraubner You are correct, I guess we must hear what Fred has to say. – Isaac Browne Dec 30 '17 at 16:39
  • I do mean it in the tetration way just to confirm. There may be an error in the title but for all purposes, it should be taken as a tetration. – Fred Wieser Dec 30 '17 at 16:40
up vote 5 down vote accepted

Let $$y=(\ln x)^{{(\ln x)}^{(\ln x)}}$$ Then $$\ln y=(\ln x)^{(\ln x)}\ln(\ln x)$$ and $$\ln(\ln y)=(\ln x)\ln(\ln x)+\ln(\ln(\ln x))$$

Differentiating, we have $$\frac1{\ln y}\cdot\frac1y\cdot\frac{dy}{dx}=\left(\frac1x\ln(\ln x)+(\ln x)\cdot\frac1{\ln x}\cdot\frac1x\right)+\frac1{\ln(\ln x)}\cdot\frac1{\ln x}\cdot \frac1x$$ so $$\frac1{y\ln y}\cdot \frac{dy}{dx}=\frac1x\left(\ln(\ln x)+1+\frac1{(\ln x)\ln(\ln x)}\right)$$ Hence $$\boxed{\frac{dy}{dx}=\frac{(\ln x)^{{(\ln x)}^{(\ln x)}}\ln\left((\ln x)^{{(\ln x)}^{(\ln x)}}\right)}x\left(\ln(\ln x)+1+\frac1{(\ln x)\ln(\ln x)}\right)}$$

Just like there is product rule and sum rule, there is also an "exponentiation rule": $$ \frac{d}{dx} f(x)^{g(x)} = g(x) f(x)^{g(x) - 1} f'(x) + f(x)^{g(x)} \ln(f(x)) g'(x) \tag{1} $$ Actually, product rule, sum rule, quotient rule, and this are all really the same theorem -- the general fact is that you can take the derivative of any expression by holding one part fixed at a time, and then adding all the results.

You can also derive the rule in (1) by taking the natural log of both sides and doing implicit differentiation: \begin{align*} y = f(x)^{g(x)} &\implies \ln y = g(x) \ln f(x) \\ &\implies \frac{1}{y} y' = g(x) \frac{f'(x)}{f(x)} + g'(x) \ln f(x) \\ &\implies y' = f(x)^{g(x)} g(x) f'(x)\frac{1}{f(x)} + f(x)^{g(x)} g'(x) \ln f(x) \\ &\implies y' = g(x) f(x)^{g(x) - 1} f'(x) + f(x)^{g(x)} \ln(f(x)) g'(x). \end{align*}

You can use this "exponentiation rule" in (1) to derive your answer.

$$\left(\left(\ln{x}\right)^{\left(\ln{x}\right)^{\ln{x}}}\right)'=\left(e^{\left(\ln{x}\right)^{\ln{x}}\ln\ln{x}}\right)'=\left(\ln{x}\right)^{\left(\ln{x}\right)^{\ln{x}}}\left(e^{\ln{x}\ln\ln{x}}\ln\ln{x}\right)'=$$ $$=\left(\ln{x}\right)^{\left(\ln{x}\right)^{\ln{x}}}\left(\ln{x}\right)^{\ln{x}}\left(\left(\ln{x}\ln\ln{x}\right)'\ln\ln{x}+\left(\ln\ln{x}\right)'\right)=$$ $$=\left(\ln{x}\right)^{\left(\ln{x}\right)^{\ln{x}}}\left(\ln{x}\right)^{\ln{x}}\left(\frac{\ln\ln{x}+1)\ln\ln{x}}{x}+\frac{1}{x\ln{x}}\right).$$

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