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Question Number 169836 by Mastermind last updated on 10/May/22

Answered by Nunzio last updated on 10/May/22

$$$$$$y=\mid x^{\mid x\mid }\mid D:x>0\Rightarrow \mid x^{\mid x\mid }\mid =x^x$$$$$$$$y=x^x$$$$\ln y=x\ln x$$$$usingimplicitdiff:$$$$\frac{1}{y}\cdot \frac{dy}{dx}=1\cdot \ln x+x\cdot \frac{1}{x}$$$$\frac{dy}{dx}=y\cdot (\ln x+1)$$$$\frac{dy}{dx}=x^x\cdot (\ln x+1)$$$$$$$$\frac{d^{2}y}{{dx}^2}=\frac{dy}{dx}[x^x\cdot (\ln x+1)]$$$$...=\frac{dy}{dx}\left(x^x\right)\cdot (\ln x+1)+x^x\cdot \frac{dy}{dx}(\ln x+1)$$$$...=x^x\cdot (\ln x+1)+x^x\cdot \frac{1}{x}$$$$...=x^x(\ln x+\frac{1}{x}+1)$$$$...=x^x\left(\frac{x\ln x+1+x}{x}\right)$$$$...=x^x\left(\frac{\ln x^x+x+1}{x}\right)$$$$...=\frac{x^x}{x}\cdot (\ln x^x+x+1)$$$$\frac{d^{2}y}{{dx}^2}=x^{x-1}\cdot (\ln x^x+x+1)$$

Commented by Mastermind last updated on 12/May/22

$$Thanks$$

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