Show that `(logx)/x` has a maximum value at `x=e` .

Explanation

Given function: `f(x) = logx / x`


Differentiating with respect to x


`f'(x)=\frac{x(1/x)-logx}
{x^2}=\frac{1-logx}{x^2}`


Now, f'(x) = 0


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