Saturday, June 23, 2007

2007/002) Find the maximum of x^(1/x)

let y = x^(1/x)
ln y = ln x/x
x ln y = ln x
differentiate both sides
x/y dy/dx + ln y = 1/x

dy/dx = (1/x - ln y)/(x/y)
= (1/x-1/x ln x)/(x/y) = y/x^2(1-ln x)
dy/dx = 0 when 1- ln x = 0
this is maximum or minimum

or x = e

for x < e dy/dx >0 or y increases

x > e dy/dx < 0 or y decreases
so x^(1/x) is maximum at x = e

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