Question #2069

If& f(x)=(1/x)[sup]x[/sup] , find f ``(1).

Expert's answer

<img src="/cgi-bin/mimetex.cgi?f(x) = (\frac{1}{x})^x = x^{-x}=e^{\ln{x^{-x}}}=e^{-x\ln{x}} \\ f'(x) = e^{-x\ln{x}}(-x\cdot1/x - \ln{x}) = e^{-x\ln{x}} (-1-\ln{x}) \\ f''(x) = e^{-x\ln{x}} (-1-\ln{x})^2 + e^{-x\ln{x}} (-1/x) = \\ =e^{-x\ln{x}}(1 + \ln^2{x}+ 2 \ln{x}- 1/x) \\ f''(1) = e^{-1\ln{1}}(1 + \ln^2{1}+ 2 \ln{1}- 1/1) = 0" title="f(x) = (\frac{1}{x})^x = x^{-x}=e^{\ln{x^{-x}}}=e^{-x\ln{x}} \\ f'(x) = e^{-x\ln{x}}(-x\cdot1/x - \ln{x}) = e^{-x\ln{x}} (-1-\ln{x}) \\ f''(x) = e^{-x\ln{x}} (-1-\ln{x})^2 + e^{-x\ln{x}} (-1/x) = \\ =e^{-x\ln{x}}(1 + \ln^2{x}+ 2 \ln{x}- 1/x) \\ f''(1) = e^{-1\ln{1}}(1 + \ln^2{1}+ 2 \ln{1}- 1/1) = 0" />

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