If $y=\frac{\log x}{x}$, then $\frac{d^2 y}{d x^2}$ at $x=1$

Question

Aicharya · Accepted Answer

$y=\frac{\log x}{x}$ $\Rightarrow \quad x y=\log x$ $\Rightarrow \quad x \frac{d y}{d x}+y=\frac{1}{x}$...(i) $\Rightarrow \quad \frac{d y}{d x}=\left(\frac{1}{x}-yight)\left(\frac{1}{x}ight)$ From Eq. (i), $x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=-\frac{1}{x^2}$ $\begin{aligned} & \Rightarrow \quad x \frac{d^2 y}{d x^2}+2\left(\frac{1}{x}-yight)\left(\frac{1}{x}ight)=-\frac{1}{x^2} \ & 	ext { At } \quad x=1,1\left(\frac{d^2 y}{d x^2}ight)+2\left(\frac{1}{1}-\frac{\log 1}{1}ight)\left(\frac{1}{1}ight)=-\frac{1}{12} \ & \Rightarrow \quad \frac{d^2 y}{d x^2}=-3\end{aligned}$

If $y=\frac{\log x}{x}$, then $\frac{d^2 y}{d x^2}$ at $x=1$

Solution