Question Number 169836 by Mastermind last updated on 10/May/22
Answered by Nunzio last updated on 10/May/22
![y=∣x^(∣x∣) ∣ D:x>0 ⇒∣x^(∣x∣) ∣=x^x y=x^x ln y=xln x using implicit diff: (1/y)∙(dy/dx)=1∙ln x+x∙(1/x) (dy/dx)=y∙(ln x+1) (dy/dx)=x^x ∙(ln x+1) (d^2 y/dx^2 ) = (dy/dx)[x^x ∙(ln x+1)] ... = (dy/dx)(x^x )∙(ln x+1)+x^x ∙(dy/dx)(ln x+1) ... = x^x ∙(ln x+1)+x^x ∙(1/x) ... = x^x (ln x+(1/x)+1) ... = x^x (((xln x+1+x)/x)) ... = x^x (((ln x^x +x+1)/x)) ... = (x^x /x)∙(ln x^x +x+1) (d^2 y/dx^2 ) = x^(x−1) ∙(ln x^x +x+1)](https://www.tinkutara.com/question/Q169840.png)
$$ \\ $$$${y}=\mid{x}^{\mid{x}\mid} \mid\:\:\:\:\:\:\:\:\:{D}:{x}>\mathrm{0}\:\Rightarrow\mid{x}^{\mid{x}\mid} \mid={x}^{{x}} \\ $$$$ \\ $$$${y}={x}^{{x}} \: \\ $$$$\mathrm{ln}\:{y}={x}\mathrm{ln}\:{x} \\ $$$${using}\:{implicit}\:{diff}: \\ $$$$\frac{\mathrm{1}}{{y}}\centerdot\frac{{dy}}{{dx}}=\mathrm{1}\centerdot\mathrm{ln}\:{x}+{x}\centerdot\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{dy}}{{dx}}={y}\centerdot\left(\mathrm{ln}\:{x}+\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}={x}^{{x}} \centerdot\left(\mathrm{ln}\:{x}+\mathrm{1}\right) \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{dy}}{{dx}}\left[{x}^{{x}} \centerdot\left(\mathrm{ln}\:{x}+\mathrm{1}\right)\right] \\ $$$$…\:=\:\frac{{dy}}{{dx}}\left({x}^{{x}} \right)\centerdot\left(\mathrm{ln}\:{x}+\mathrm{1}\right)+{x}^{{x}} \centerdot\frac{{dy}}{{dx}}\left(\mathrm{ln}\:{x}+\mathrm{1}\right) \\ $$$$…\:=\:{x}^{{x}} \centerdot\left(\mathrm{ln}\:{x}+\mathrm{1}\right)+{x}^{{x}} \centerdot\frac{\mathrm{1}}{{x}} \\ $$$$…\:=\:{x}^{{x}} \left(\mathrm{ln}\:{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right) \\ $$$$\:…\:=\:{x}^{{x}} \left(\frac{{x}\mathrm{ln}\:{x}+\mathrm{1}+{x}}{{x}}\right) \\ $$$$…\:=\:{x}^{{x}} \left(\frac{\mathrm{ln}\:{x}^{{x}} +{x}+\mathrm{1}}{{x}}\right) \\ $$$$…\:=\:\frac{{x}^{{x}} }{{x}}\centerdot\left(\mathrm{ln}\:{x}^{{x}} +{x}+\mathrm{1}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:{x}^{{x}−\mathrm{1}} \centerdot\left(\mathrm{ln}\:{x}^{{x}} +{x}+\mathrm{1}\right) \\ $$
Commented by Mastermind last updated on 12/May/22
$${Thanks} \\ $$