Question Number 66046 by olalekan2 last updated on 08/Aug/19
$${find}\:\frac{{dy}}{{dx}}\:{if}\:{y}={x}^{{x}^{{x}} } \\ $$$${help}\:{pls} \\ $$
Commented by Prithwish sen last updated on 08/Aug/19
![y=x^x^x ⇒lny=x^x lnx let u=x^x ⇒lnu=xlnx⇒(du/dx) = u(lnx+1)=x^x (lnx+1) lny=ulnx⇒(1/y) (dy/dx) =(du/dx)lnx +(u/x) (dy/dx) = x^x^x [x^x (lnx+1)lnx +(x^x /x)] please check.](https://www.tinkutara.com/question/Q66047.png)
$$\mathrm{y}=\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \Rightarrow\mathrm{lny}=\mathrm{x}^{\mathrm{x}} \mathrm{lnx} \\ $$$$\mathrm{let}\:\mathrm{u}=\mathrm{x}^{\mathrm{x}} \Rightarrow\mathrm{lnu}=\mathrm{xlnx}\Rightarrow\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{u}\left(\mathrm{lnx}+\mathrm{1}\right)=\mathrm{x}^{\mathrm{x}} \left(\mathrm{lnx}+\mathrm{1}\right) \\ $$$$\mathrm{lny}=\mathrm{ulnx}\Rightarrow\frac{\mathrm{1}}{\mathrm{y}}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\frac{\mathrm{du}}{\mathrm{dx}}\mathrm{lnx}\:+\frac{\mathrm{u}}{\mathrm{x}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \left[\mathrm{x}^{\mathrm{x}} \left(\mathrm{lnx}+\mathrm{1}\right)\mathrm{lnx}\:+\frac{\mathrm{x}^{\mathrm{x}} }{\mathrm{x}}\right]\: \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$
Commented by olalekan2 last updated on 17/Oct/19
$${thanks}…….. \\ $$