Show that `f(x)=(log x)/(x), x ne 0` is maximum at `x = e`.

Explanation

`f(x)=(log x)/(x)`
`:.f'(x)=("d")/("dx")((logx)/(x))`
`=(x("d")/("dx")(logx)-logx("d")/("dx")(x))/(x^(2))`
`=(x((1)/(x))-(logx)(1))/(x^(2))=(1-logx)/(x^(2))" and"`
`f''(x)=("d")/("dx")((1-logx)/(x^(2)))`
`=(x^(2)("d")/("dx")(1-logx)-(1-logx)("d")/("dx")(x^(2)))/(x^(4))`
`=(x^(2)(0-(1)/(x))-(1-logx)xx2x)/(x^(4))`
`=(-x-2x+2xlogx)/(x^(4))=(x(2 logx-3))/(x^(4))`
`:.f^(n)(x)=(2 log x-3)/(x^(3))`
Now `f'(x)=0`, if `(1-log x)/(x^(2))=0`
i.e. if `1-logx=0`, i.e. if `logx=1=log e`
i.e. if x = e
and `f^(n)(e)=(2loge-3)/(e^(3))=(-1)/(e^(3))lt 0" "...[because log e=1]`
`:.` by the second derivative test, f(x) is maximum at x = e.