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Question-26447




Question Number 26447 by yesaditya22@gmail.com last updated on 25/Dec/17
Commented by kaivan.ahmadi last updated on 25/Dec/17
y=x^x^x  ⇒lny=x^x lnx⇒(y^′ /y)=(x^x )^′ lnx+(x^x /x)⇒  y^′ =y[(x^x )^′ lnx+x^(x−1) ]  we compute (x^x )^′ , then replace in below relation  g=x^x ⇒lng=xlnx⇒(g^′ /g)=lnx+1⇒  g^′ =x^x (lnx+1)
$$\mathrm{y}=\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \Rightarrow\mathrm{lny}=\mathrm{x}^{\mathrm{x}} \mathrm{lnx}\Rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}}=\left(\mathrm{x}^{\mathrm{x}} \right)^{'} \mathrm{lnx}+\frac{\mathrm{x}^{\mathrm{x}} }{\mathrm{x}}\Rightarrow \\ $$$$\mathrm{y}^{'} =\mathrm{y}\left[\left(\mathrm{x}^{\mathrm{x}} \right)^{'} \mathrm{lnx}+\mathrm{x}^{\mathrm{x}−\mathrm{1}} \right] \\ $$$$\mathrm{we}\:\mathrm{compute}\:\left(\mathrm{x}^{\mathrm{x}} \right)^{'} ,\:\mathrm{then}\:\mathrm{replace}\:\mathrm{in}\:\mathrm{below}\:\mathrm{relation} \\ $$$$\mathrm{g}=\mathrm{x}^{\mathrm{x}} \Rightarrow\mathrm{lng}=\mathrm{xlnx}\Rightarrow\frac{\mathrm{g}^{'} }{\mathrm{g}}=\mathrm{lnx}+\mathrm{1}\Rightarrow \\ $$$$\mathrm{g}^{'} =\mathrm{x}^{\mathrm{x}} \left(\mathrm{lnx}+\mathrm{1}\right) \\ $$

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