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Question Number 120243 by benjo_mathlover last updated on 30/Oct/20

Answered by bemath last updated on 30/Oct/20

$$letf\left(x\right)=x^{\frac{1}{x^6}}\iff \ln f\left(x\right)=\frac{\ln \left(x\right)}{x^6}$$$$differentiatingbothsidesgives$$$$\frac{f{}^{\prime }\left(x\right)}{f\left(x\right)}=\frac{x^5-6x^5\ln \left(x\right)}{x^{12}};f{}^{\prime }\left(x\right)=\frac{x^{\frac{1}{x^6}}.x^5(1-6\ln \left(x\right))}{x^{12}}$$$$f{}^{\prime }\left(x\right)=\frac{x^{\frac{1}{x^6}}(1-6\ln x)}{x^7}$$$$takingf{}^{\prime }\left(x\right)=0,weget\ln x=\frac{1}{6}orx=e^{\frac{1}{6}}$$$$curveincreasingon0<x<e^{\frac{1}{6}}and$$$$decreasingonx>e^{\frac{1}{6}}.Thusmaximum$$$$valueis{\left(e^{\frac{1}{6}}\right)}^{\frac{1}{e}}=e^{\frac{1}{6e}}$$

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