If y=sin (log _e x),then x^2 frac{d^2 y}{d x^ | vedclass

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(D) Given that,$y=\sin (\log _e x)$ $(i)$. On differentiating with respect to $x$,we get: $\frac{d y}{d x} = \cos (\log _e x) \cdot \frac{1}{x} \impli

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$-y$

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(D) Given that,$y=\sin (\log _e x)$ $(i)$. On differentiating with respect to $x$,we get: $\frac{d y}{d x} = \cos (\log _e x) \cdot \frac{1}{x} \implies x \frac{d y}{d x} = \cos (\log _e x)$. Again differentiating with respect to $x$ using the product rule on the left side: $x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = -\sin (\log _e x) \cdot \frac{1}{x}$. Multiplying both sides by $x$: $x^2 \frac{d^2 y}{d x^2} + x \frac{d y}{d x} = -\sin (\log _e x)$. Since $y = \sin (\log _e x)$,we have: $x^2 \frac{d^2 y}{d x^2} + x \frac{d y}{d x} = -y$.

If $y=\sin (\log _e x)$,then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to

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