If cos ^{-1}left(frac{y}{b}right)=2 log left( | vedclass

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(B) Given: $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$ for $x>0$. On differentiating with respect to $x$, we get: $-\frac{1}{

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$-4 y$

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(B) Given: $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$ for $x>0$. On differentiating with respect to $x$, we get: $-\frac{1}{\sqrt{1-\frac{y^2}{b^2}}} \cdot \frac{1}{b} \frac{d y}{d x} = 2 \cdot \frac{1}{\left(\frac{x}{2}\right)} \cdot \frac{1}{2}$ $-\frac{1}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x} = \frac{2}{x}$ $x \frac{d y}{d x} = -2 \sqrt{b^2-y^2} \quad \dots (i)$ Again, differentiating with respect to $x$: $x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = -2 \cdot \frac{1}{2} (b^2-y^2)^{-1/2} (-2y) \frac{d y}{d x}$ $x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{2y}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x}$ From equation $(i)$, we have $\sqrt{b^2-y^2} = -\frac{x}{2} \frac{d y}{d x}$. Substituting this: $x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{2y}{-\frac{x}{2} \frac{d y}{d x}} \cdot \frac{d y}{d x}$ $x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = -\frac{4y}{x}$ Multiplying by $x$: $x^2 \frac{d^2 y}{d x^2} + x \frac{d y}{d x} = -4y$.

If $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$, where $x>0$, then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to

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