Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=\frac{\log x}{x}$, then $\frac{d^2 y}{d x^2}$ at $x=1$
Options:
Solution:
1800 Upvotes
Verified Answer
The correct answer is:
$-3$
$y=\frac{\log x}{x}$
$\Rightarrow \quad x y=\log x$
$\Rightarrow \quad x \frac{d y}{d x}+y=\frac{1}{x}$...(i)
$\Rightarrow \quad \frac{d y}{d x}=\left(\frac{1}{x}-y\right)\left(\frac{1}{x}\right)$
From Eq. (i), $x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=-\frac{1}{x^2}$
$\begin{aligned} & \Rightarrow \quad x \frac{d^2 y}{d x^2}+2\left(\frac{1}{x}-y\right)\left(\frac{1}{x}\right)=-\frac{1}{x^2} \\ & \text { At } \quad x=1,1\left(\frac{d^2 y}{d x^2}\right)+2\left(\frac{1}{1}-\frac{\log 1}{1}\right)\left(\frac{1}{1}\right)=-\frac{1}{12} \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=-3\end{aligned}$
$\Rightarrow \quad x y=\log x$
$\Rightarrow \quad x \frac{d y}{d x}+y=\frac{1}{x}$...(i)
$\Rightarrow \quad \frac{d y}{d x}=\left(\frac{1}{x}-y\right)\left(\frac{1}{x}\right)$
From Eq. (i), $x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=-\frac{1}{x^2}$
$\begin{aligned} & \Rightarrow \quad x \frac{d^2 y}{d x^2}+2\left(\frac{1}{x}-y\right)\left(\frac{1}{x}\right)=-\frac{1}{x^2} \\ & \text { At } \quad x=1,1\left(\frac{d^2 y}{d x^2}\right)+2\left(\frac{1}{1}-\frac{\log 1}{1}\right)\left(\frac{1}{1}\right)=-\frac{1}{12} \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=-3\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.