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!