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Question Number 48307 by MJS last updated on 21/Nov/18

$$f\left(x\right)=x^{x^{x-1}}[=x^{\left(x^{x-1}\right)}]$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{funny}\mathrm{one}...$$$$f\left(1\right)=1$$$$f\left(2\right)=2^2$$$$f\left(3\right)={\left(3^3\right)}^3$$$$f\left(4\right)={\left({\left(4^4\right)}^4\right)}^4$$$$f\left(5\right)={\left({\left({\left(5^5\right)}^5\right)}^5\right)}^5$$$$...$$$$\mathrm{find}{f}^{\prime }\left(x\right)$$

Commented by mr W last updated on 22/Nov/18

$$g\left(x\right)=x^{x-1}=e^{(x-1)\ln x}$$$$g^{\prime }\left(x\right)=e^{(x-1)\ln x}[\frac{x-1}{x}+\ln x]=x^{x-1}(1+\ln x-\frac{1}{x})$$$$f\left(x\right)=x^{g\left(x\right)}=e^{g\left(x\right)\ln x}$$$$f^{\prime }\left(x\right)=e^{g\left(x\right)\ln x}[\frac{g\left(x\right)}{x}+\left(\ln x\right)g^{\prime }\left(x\right)]$$$$f^{\prime }\left(x\right)=x^{x^{x-1}}[\frac{x^{x-1}}{x}+\left(\ln x\right)x^{x-1}(1+\ln x-\frac{1}{x})]$$$$\Rightarrow f^{\prime }\left(x\right)=x^{x^{x-1}}{x}^{x-1}[\frac{1-\ln x}{x}+\ln x+{\left(\ln x\right)}^2]$$

Commented by MJS last updated on 22/Nov/18

$$\mathrm{thank}\mathrm{you}$$

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