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what is the max value of y=x^22/10x^logx

  • Maths - ,

    y = x^22 / 10x^logx
    Nasty, but straightforward
    Recall that d/dx (u^v) = v/u u^v u' + lnu u^v v'

    I will assume by logx you mean lnx. If not, just recall that logx = lnx/ln10

    d/dx x^lnx = (2/x lnx) x^lnx

    So, using the quotient rule, we have

    y' = 22x^21/(10x^logx) + x^22(20/x logx) x^logx
    = 1/5 x^21/x^logx (11 - logx)

    y'=0 when logx = 11, or x = e^11

    y(e^11) = 1/10 e^121

    whew - that's big!

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