Posted by **Tony** on Sunday, June 30, 2013 at 3:36pm.

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

- Maths -
**Steve**, Sunday, June 30, 2013 at 5:52pm
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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