Question

If $y=\frac{\log x}{x}$ then $y_2\left(e\right)$ equals

• A. $\frac{1}{e^3}$
• B. $\frac{1}{e^2}$
• C. $-\frac{1}{e^2}$
• D. $-\frac{1}{e^3}$

Answer 1

The correct option is D $-\frac{1}{e^3}$

$y=\frac{\log x}{x}$
Differentiating it w.r.t to x
$y^{\prime }=\frac{1-\log x}{x^2}$
Again differentiating it w.r.t. x
$y^{\prime \prime }=\frac{x^2(-\frac{1}{x})-(1-\log x)2x}{x^4}=\frac{-x-2x(1-\log x)}{x^4}=\frac{-x(1+2(1-\log x))}{x^4}$
Hence
$y^{\prime \prime }\left(e\right)=-\frac{1+2(1-\log e)}{e^3}=\frac{-1}{e^3}$