Question

y=lnx÷x ,x > 0 is increasing

Answer 1

#Hey there!!
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given that:
.
.=) y = lnx/x

=> differentiate it w.r .t x we get,

=> y' = (x.1/x - lnx)/x²

=> y' = (1 - lnx)/x²


now for increasing function , y'>0

=> (1-lnx)/x² > 0

=> (1 - lnx) > 0

=> -lnx > - 1

=> lnx < 1

◆ since lnx is decreasing functions so it will change the equality sign.

=> lnx < ln0

=> x > 0

so for all values of x >0 , y is increasing.

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◆ Hope it will help you