Question Number 48307 by MJS last updated on 21/Nov/18
![f(x)=x^x^(x−1) [=x^((x^(x−1) )) ] it′s a funny one... f(1)=1 f(2)=2^2 f(3)=(3^3 )^3 f(4)=((4^4 )^4 )^4 f(5)=(((5^5 )^5 )^5 )^5 ... find f′(x)](https://www.tinkutara.com/question/Q48307.png)
$${f}\left({x}\right)={x}^{{x}^{{x}−\mathrm{1}} } \:\:\:\:\:\left[={x}^{\left({x}^{{x}−\mathrm{1}} \right)} \right] \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{funny}\:\mathrm{one}… \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{2}} \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{3}} \\ $$$${f}\left(\mathrm{4}\right)=\left(\left(\mathrm{4}^{\mathrm{4}} \right)^{\mathrm{4}} \right)^{\mathrm{4}} \\ $$$${f}\left(\mathrm{5}\right)=\left(\left(\left(\mathrm{5}^{\mathrm{5}} \right)^{\mathrm{5}} \right)^{\mathrm{5}} \right)^{\mathrm{5}} \\ $$$$… \\ $$$$\mathrm{find}\:{f}'\left({x}\right) \\ $$
Commented by mr W last updated on 22/Nov/18
![g(x)=x^(x−1) =e^((x−1)ln x) g′(x)=e^((x−1)ln x) [((x−1)/x)+ln x]=x^(x−1) (1+ln x−(1/x)) f(x)=x^(g(x)) =e^(g(x) ln x ) f′(x)=e^(g(x) ln x) [((g(x))/x)+(ln x)g′(x)] f′(x)=x^x^(x−1) [(x^(x−1) /x)+(ln x)x^(x−1) (1+ln x−(1/x))] ⇒f′(x)=x^x^(x−1) x^(x−1) [((1−ln x)/x)+ln x+(ln x)^2 ]](https://www.tinkutara.com/question/Q48313.png)
$${g}\left({x}\right)={x}^{{x}−\mathrm{1}} ={e}^{\left({x}−\mathrm{1}\right)\mathrm{ln}\:{x}} \\ $$$${g}'\left({x}\right)={e}^{\left({x}−\mathrm{1}\right)\mathrm{ln}\:{x}} \left[\frac{{x}−\mathrm{1}}{{x}}+\mathrm{ln}\:{x}\right]={x}^{{x}−\mathrm{1}} \left(\mathrm{1}+\mathrm{ln}\:{x}−\frac{\mathrm{1}}{{x}}\right) \\ $$$${f}\left({x}\right)={x}^{{g}\left({x}\right)} ={e}^{{g}\left({x}\right)\:\mathrm{ln}\:{x}\:} \\ $$$${f}'\left({x}\right)={e}^{{g}\left({x}\right)\:\mathrm{ln}\:{x}} \left[\frac{{g}\left({x}\right)}{{x}}+\left(\mathrm{ln}\:{x}\right){g}'\left({x}\right)\right] \\ $$$${f}'\left({x}\right)={x}^{{x}^{{x}−\mathrm{1}} } \left[\frac{{x}^{{x}−\mathrm{1}} }{{x}}+\left(\mathrm{ln}\:{x}\right){x}^{{x}−\mathrm{1}} \left(\mathrm{1}+\mathrm{ln}\:{x}−\frac{\mathrm{1}}{{x}}\right)\right] \\ $$$$\Rightarrow{f}'\left({x}\right)={x}^{{x}^{{x}−\mathrm{1}} } {x}^{{x}−\mathrm{1}} \left[\frac{\mathrm{1}−\mathrm{ln}\:{x}}{{x}}+\mathrm{ln}\:{x}+\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \right] \\ $$
Commented by MJS last updated on 22/Nov/18
$$\mathrm{thank}\:\mathrm{you} \\ $$