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Question Number 77859 by jagoll last updated on 11/Jan/20

$$y=\mid x\mid {}^{\mid x\mid }$$$$\frac{dy}{dx}=?$$

Commented by mathmax by abdo last updated on 11/Jan/20

$$y\left(x\right)=e^{\mid x\mid ln\mid x\mid }ifx>0\Rightarrow y\left(x\right)=e^{xln\left(x\right)}\Rightarrow y^{{}^{\prime }}\left(x\right)={\left(xlnx\right)}^{{}^{\prime }}{e}^{xlnx}$$$$=(lnx+1)x^x$$$$ifx<0\Rightarrow y\left(x\right)=e^{-xln(-x)}\Rightarrow y^{{}^{\prime }}\left(x\right)={(-xln(-x))}^{{}^{\prime }}{e}^{-xln(-x)}$$$$=(-ln(-x)-x\times \frac{-1}{-x}){(-x)}^{-x}=(-x-ln(-x)){(-x)}^{-x}$$

Answered by mr W last updated on 11/Jan/20

$$ifx>0:$$$$y=x^x$$$$\ln y=x\ln x$$$$\frac{y^{\prime }}{y}=\ln x+1$$$$y^{\prime }=(1+\ln x)x^x$$$$$$$$ifx<0:$$$$y={(-x)}^{(-x)}$$$$\ln y=-x\ln (-x)$$$$\frac{y^{\prime }}{y}=-\ln (-x)-1$$$$y^{\prime }=-(1+\ln (-x)){(-x)}^{(-x)}$$$$$$$$\Rightarrow y^{\prime }=sign\left(x\right)(1+\ln \mid x\mid )\mid x{\mid }^{\mid x\mid }$$

Commented by jagoll last updated on 11/Jan/20

$$whatissign\left(x\right)sir?$$

Commented by mr W last updated on 11/Jan/20

$$sign\left(x\right)=\{\begin{array}{l}1,ifx>0\\ 0,ifx=0\\ -1,ifx<0\end{array}$$

Answered by john santu last updated on 11/Jan/20

$$y=e^{\mid x\mid ln(\mid x\mid )}\Rightarrow \frac{dy}{dx}=e^{\mid x\mid ln(\mid x\mid )}(\frac{x}{\mid x\mid }ln(\mid x\mid )+\mid x\mid \frac{1}{\mid x\mid }\frac{x}{\mid x\mid })$$$$\frac{dy}{dx}=\mid x{\mid }^{\mid x\mid }(\frac{x}{\mid x\mid }ln(\mid x\mid )+\frac{x}{\mid x\mid })$$

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