Question

If $y=\frac{lnx}{x}$, then the value of $y^{\prime \prime }(e)$ is

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

Answer 1

Answer: (D)

$\because y=\frac{lnx}{x}$
On differentiating $w.r.t.x$, we get
$y^{\prime }=\frac{\left(1-lnx\right)}{x^2}$
$\Rightarrow y^{\prime \prime }=\frac{x^2\left(-\frac{1}{x}\right)-\left(1-lnx\right)2x}{x^4}$
$\therefore y^{\prime \prime }\left(e\right)=\frac{-e-0}{e^4}$
$=-\frac{1}{e^3}$