Q105 / 160 · AP EAMCET 2008 · Mathematics
If $y=\sin \left(\log _e x\right)$, then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to
- A$\sin \left(\log _e x\right)$
- B$\cos \left(\log _e x\right)$
- C$y^2$
- Correct answerD$-y$
Solution
Given that,
$$
y=\sin \left(\log _e x\right)
$$
On differentiating w.r.t. $x$, we get
$$
\frac{d y}{d x}=\cos \left(\log _e x\right) \cdot \frac{1}{x}
$$
Again differentiating, we get
$$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{-x \cdot \sin \left(\log _e x\right) \cdot \frac{1}{x}-\cos \left(\log _e x\right) \cdot 1}{x^2} \\
& =\frac{-\sin \left(\log _e x\right)-\cos \left(\log _e x\right)}{x^2}
\end{aligned}
$$
Now, $\quad x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$
$$
\begin{aligned}
& =\frac{x^2\left[-\sin \left(\log _e x\right)-\cos \left(\log _e x\right)\right]}{x^2} \\
& =-\sin \left(\log _e x\right) \\
& =-y \quad \text { [from Eq. (i)] }
\end{aligned}
$$
Hence, the correct answer is option: 'D'.