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Q.
If $ y=\frac{ln\,x}{x}$, then the value of $y''(e)$ is
Continuity and Differentiability
Solution:
$\because y=\frac{ln\,x}{x}$
On differentiating $w.r.t. x$, we get
$y'=\frac{\left(1-ln\,x\right)}{x^{2}}$
$\Rightarrow y''=\frac{x^{2}\left(-\frac{1}{x}\right)-\left(1-ln\,x\right)2x}{x^{4}}$
$\therefore y''\left(e\right)=\frac{-e-0}{e^{4}}$
$=-\frac{1}{e^{3}}$