Question
If $\text{y}=\frac{\log\text{x}}{\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\log\text{x}-3}{\text{x}^3}.$

Answer

Here,
$\text{y}=\frac{\log\text{x}}{\text{x}},$
Differentiating w.r.t.x, we get
$\frac{\text{d}\text{y}}{\text{dx}}=\frac{1-\log\text{x}}{\text{x}^2}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{x}-2\text{x}(1-\log\text{x})}{\text{x}^4}$
$=\frac{-\text{x}-2\text{x}+2\text{x}\log\text{x}}{\text{x}^4}$
$=\frac{-3+2\log\text{x}}{\text{x}^3}$
$=\frac{2\log\text{x}-3}{\text{x}^3}$
Hence proved

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