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Question Number 5365 by sanusihammed last updated on 11/May/16

If  y = x^x^x^x      .  Find dy/dx

$${If}\:\:{y}\:=\:{x}^{{x}^{{x}^{{x}} } } \:\:\:.\:\:{Find}\:{dy}/{dx} \\ $$

Answered by Yozzii last updated on 11/May/16

lny=x^x^x  lnx   (x>0)  By implicit differentiation we get  y^(−1) y^′ =(1/x)x^x^x  +(d/dx)(x^x^x  )lnx  ⇒(dy/dx)=y(x^(x^x −1) +(d/dx)(x^x^x  )lnx)=x^x^x^x   (x^(x^x −1) +(d/dx)(x^x^x  )lnx)    Let g=x^x^x   where x>0.  ⇒lng=x^x lnx  ⇒g^(−1) g^′ =x^(x−1) +(d/dx)(x^x )lnx  (d/dx)(x^x^x  )=g^′ =g(x^(x−1) +(d/dx)(x^x )lnx)=x^x^x  (x^(x−1) +(d/dx)(x^x )lnx)    Let r=x^x   (x>0).  ⇒lnr=xlnx  ⇒r^(−1) r^′ =lnx+1⇒r′=r(1+lnx)=x^x (1+lnx)=(d/dx)(x^x )  ∴(d/dx)(x^x^x  )=x^x^x  (x^(x−1) +x^x (1+lnx)lnx)    ∴(dy/dx)=x^x^x^x   (x^(x^x −1) +x^x^x  (x^(x−1) +x^x (1+lnx)lnx)lnx)  (dy/dx)=x^(x^x^x  +x^x −1) +x^(x^x^x  +x^x ) (x^(x−1) +x^x (lnx+ln^2 x))lnx  (dy/dx)=x^(x^x^x  +x^x −1) +x^(x^x^x  +x^x +x−1) lnx+x^(x^x^x  +x^x +x) ln^2 x+x^(x^x^x  +x^x +x) ln^3 x  (x>0)

$${lny}={x}^{{x}^{{x}} } {lnx}\:\:\:\left({x}>\mathrm{0}\right) \\ $$$${By}\:{implicit}\:{differentiation}\:{we}\:{get} \\ $$$${y}^{−\mathrm{1}} {y}^{'} =\frac{\mathrm{1}}{{x}}{x}^{{x}^{{x}} } +\frac{{d}}{{dx}}\left({x}^{{x}^{{x}} } \right){lnx} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={y}\left({x}^{{x}^{{x}} −\mathrm{1}} +\frac{{d}}{{dx}}\left({x}^{{x}^{{x}} } \right){lnx}\right)={x}^{{x}^{{x}^{{x}} } } \left({x}^{{x}^{{x}} −\mathrm{1}} +\frac{{d}}{{dx}}\left({x}^{{x}^{{x}} } \right){lnx}\right) \\ $$$$ \\ $$$${Let}\:{g}={x}^{{x}^{{x}} } \:{where}\:{x}>\mathrm{0}. \\ $$$$\Rightarrow{lng}={x}^{{x}} {lnx} \\ $$$$\Rightarrow{g}^{−\mathrm{1}} {g}^{'} ={x}^{{x}−\mathrm{1}} +\frac{{d}}{{dx}}\left({x}^{{x}} \right){lnx} \\ $$$$\frac{{d}}{{dx}}\left({x}^{{x}^{{x}} } \right)={g}^{'} ={g}\left({x}^{{x}−\mathrm{1}} +\frac{{d}}{{dx}}\left({x}^{{x}} \right){lnx}\right)={x}^{{x}^{{x}} } \left({x}^{{x}−\mathrm{1}} +\frac{{d}}{{dx}}\left({x}^{{x}} \right){lnx}\right) \\ $$$$ \\ $$$${Let}\:{r}={x}^{{x}} \:\:\left({x}>\mathrm{0}\right). \\ $$$$\Rightarrow{lnr}={xlnx} \\ $$$$\Rightarrow{r}^{−\mathrm{1}} {r}^{'} ={lnx}+\mathrm{1}\Rightarrow{r}'={r}\left(\mathrm{1}+{lnx}\right)={x}^{{x}} \left(\mathrm{1}+{lnx}\right)=\frac{{d}}{{dx}}\left({x}^{{x}} \right) \\ $$$$\therefore\frac{{d}}{{dx}}\left({x}^{{x}^{{x}} } \right)={x}^{{x}^{{x}} } \left({x}^{{x}−\mathrm{1}} +{x}^{{x}} \left(\mathrm{1}+{lnx}\right){lnx}\right) \\ $$$$ \\ $$$$\therefore\frac{{dy}}{{dx}}={x}^{{x}^{{x}^{{x}} } } \left({x}^{{x}^{{x}} −\mathrm{1}} +{x}^{{x}^{{x}} } \left({x}^{{x}−\mathrm{1}} +{x}^{{x}} \left(\mathrm{1}+{lnx}\right){lnx}\right){lnx}\right) \\ $$$$\frac{{dy}}{{dx}}={x}^{{x}^{{x}^{{x}} } +{x}^{{x}} −\mathrm{1}} +{x}^{{x}^{{x}^{{x}} } +{x}^{{x}} } \left({x}^{{x}−\mathrm{1}} +{x}^{{x}} \left({lnx}+{ln}^{\mathrm{2}} {x}\right)\right){lnx} \\ $$$$\frac{{dy}}{{dx}}={x}^{{x}^{{x}^{{x}} } +{x}^{{x}} −\mathrm{1}} +{x}^{{x}^{{x}^{{x}} } +{x}^{{x}} +{x}−\mathrm{1}} {lnx}+{x}^{{x}^{{x}^{{x}} } +{x}^{{x}} +{x}} {ln}^{\mathrm{2}} {x}+{x}^{{x}^{{x}^{{x}} } +{x}^{{x}} +{x}} {ln}^{\mathrm{3}} {x}\:\:\left({x}>\mathrm{0}\right) \\ $$

Commented by sanusihammed last updated on 12/May/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

Commented by sanusihammed last updated on 12/May/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

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