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
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
- $\sin \left(\log _e x\right)$
- $\cos \left(\log _e x\right)$
- $y^2$
- $-y$
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, $\quad x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$
$\begin{aligned}
& =\frac{x^2\left[-\sin \left(\log _e x\right)-\cos \left(\log _e x\right)\right]}{x^2} \\
& =-\sin \left(\log _e x\right) \\
& =-y \quad \text { [from Eq. (i)] }
\end{aligned}$
Asked in: MHT CET Full Test 12
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