Question Number 10711 by Saham last updated on 23/Feb/17
$$\mathrm{If}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:,\:\:\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$
Answered by mrW1 last updated on 23/Feb/17
![u=x^x ln u=xln x (1/u)×(du/dx)=ln x+x×(1/x)=1+ln x ⇒(du/dx)=u(1+ln x)=x^x (1+ln x) y=x^u ln y=uln x (1/y)×(dy/dx)=(du/dx)×ln x+u×(1/x)=x^x (1+ln x)ln x+x^(x−1) ⇒(dy/dx)=y×[x^x (1+ln x)ln x+x^(x−1) ] =x^x^x ×[x^x (1+ln x)ln x+x^(x−1) ] =x^(x^x +x) (1+ln x)ln x+x^(x^x +x−1) =x^(x^x +x−1) +x^(x^x +x) ln x+x^(x^x +x) (ln x)^2](https://www.tinkutara.com/question/Q10715.png)
$${u}={x}^{{x}} \\ $$$$\mathrm{ln}\:{u}={x}\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{u}}×\frac{{du}}{{dx}}=\mathrm{ln}\:{x}+{x}×\frac{\mathrm{1}}{{x}}=\mathrm{1}+\mathrm{ln}\:{x} \\ $$$$\Rightarrow\frac{{du}}{{dx}}={u}\left(\mathrm{1}+\mathrm{ln}\:{x}\right)={x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right) \\ $$$$ \\ $$$${y}={x}^{{u}} \\ $$$$\mathrm{ln}\:{y}={u}\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}×\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}×\mathrm{ln}\:{x}+{u}×\frac{\mathrm{1}}{{x}}={x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right)\mathrm{ln}\:{x}+{x}^{{x}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={y}×\left[{x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right)\mathrm{ln}\:{x}+{x}^{{x}−\mathrm{1}} \right] \\ $$$$={x}^{{x}^{{x}} } ×\left[{x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right)\mathrm{ln}\:{x}+{x}^{{x}−\mathrm{1}} \right] \\ $$$$={x}^{{x}^{{x}} +{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right)\mathrm{ln}\:{x}+{x}^{{x}^{{x}} +{x}−\mathrm{1}} \\ $$$$={x}^{{x}^{{x}} +{x}−\mathrm{1}} +{x}^{{x}^{{x}} +{x}} \mathrm{ln}\:{x}+{x}^{{x}^{{x}} +{x}} \left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \\ $$
Commented by Saham last updated on 23/Feb/17
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$