Question Number 179258 by zaheen last updated on 27/Oct/22
$${y}={x}^{{x}^{{x}} } \\ $$$$\frac{{dy}}{{dx}}=? \\ $$
Commented by CElcedricjunior last updated on 30/Oct/22
![y=x^x^x =e^(x^x ln(x)) si x∈]0;∞[ (dy/dx)=[x^x ((1/x))+lnx(e^(xlnx) )′]e^(x^x lnx) =(x^(x−1) +lnx(1+lnx)x^x )x^x^x =(x^(x−1) +x^x lnx+x^x ln^2 x)x^x^x =x^(x^x +x) ((1/x)+lnx+ln^2 x) ...................le celebre cedric junior...](https://www.tinkutara.com/question/Q179557.png)
$$\left.{y}=\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} } =\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)} \:\:\boldsymbol{\mathrm{si}}\:\boldsymbol{\mathrm{x}}\in\right]\mathrm{0};\infty\left[\right. \\ $$$$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\left[\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)+\boldsymbol{\mathrm{lnx}}\left(\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{xlnx}}} \right)'\right]\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{lnx}}} \\ $$$$\:\:\:\:\:=\left(\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} +\boldsymbol{\mathrm{lnx}}\left(\mathrm{1}+\boldsymbol{\mathrm{lnx}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\:\:\:\:\:=\left(\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} +\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{lnx}}+\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}x}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\:\:\:\:\:=\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{x}}} \left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}+\boldsymbol{\mathrm{lnx}}+\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}x}}\right) \\ $$$$……………….{le}\:{celebre}\:{cedric}\:{junior}… \\ $$
Answered by mr W last updated on 27/Oct/22
![let g(x)=x^x =e^(xln x) ((dg(x))/dx)=e^(xln x) (1+ln x)=(1+ln x)x^x y=x^(g(x)) =e^(g(x)ln x) (dy/dx)=e^(g(x)ln x) [(ln x)×((dg(x))/dx)+((g(x))/x)] =x^x^x [(ln x)(1+ln x)x^x +x^(x−1) ]](https://www.tinkutara.com/question/Q179260.png)
$${let}\:{g}\left({x}\right)={x}^{{x}} ={e}^{{x}\mathrm{ln}\:{x}} \\ $$$$\frac{{dg}\left({x}\right)}{{dx}}={e}^{{x}\mathrm{ln}\:{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right)=\left(\mathrm{1}+\mathrm{ln}\:{x}\right){x}^{{x}} \\ $$$${y}={x}^{{g}\left({x}\right)} ={e}^{{g}\left({x}\right)\mathrm{ln}\:{x}} \\ $$$$\frac{{dy}}{{dx}}={e}^{{g}\left({x}\right)\mathrm{ln}\:{x}} \left[\left(\mathrm{ln}\:{x}\right)×\frac{{dg}\left({x}\right)}{{dx}}+\frac{{g}\left({x}\right)}{{x}}\right] \\ $$$$\:\:\:\:\:={x}^{{x}^{{x}} } \left[\left(\mathrm{ln}\:{x}\right)\left(\mathrm{1}+\mathrm{ln}\:{x}\right){x}^{{x}} +{x}^{{x}−\mathrm{1}} \right] \\ $$
Commented by zaheen last updated on 27/Oct/22
$${thanks}\:{sir} \\ $$