If y = sqrt{sin x + sqrt{sin x + sqrt{sin x + | vedclass

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(B) Given $y = \sqrt{\sin x + y}$. Squaring both sides,we get $y^2 = \sin x + y$.Differentiating with respect to $x$:$2y \frac{dy}{dx} = \cos x + \fra

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$-2$

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(B) Given $y = \sqrt{\sin x + y}$. Squaring both sides,we get $y^2 = \sin x + y$.Differentiating with respect to $x$:$2y \frac{dy}{dx} = \cos x + \frac{dy}{dx} \implies \frac{dy}{dx}(2y - 1) = \cos x \implies \frac{dy}{dx} = \frac{\cos x}{2y - 1}$.Differentiating again with respect to $x$ using the quotient rule:$\frac{d^2 y}{dx^2} = \frac{(2y - 1)(-\sin x) - \cos x (2 \frac{dy}{dx})}{(2y - 1)^2}$.Substituting $\frac{dy}{dx} = \frac{\cos x}{2y - 1}$:$\frac{d^2 y}{dx^2} = \frac{-(2y - 1)\sin x - \frac{2 \cos^2 x}{2y - 1}}{(2y - 1)^2}$.At the point $(\pi, 1)$,we have $x = \pi$ and $y = 1$:$\sin \pi = 0$ and $\cos \pi = -1$.$\left(\frac{d^2 y}{dx^2}\right)_{(\pi, 1)} = \frac{-(2(1) - 1)(0) - \frac{2(-1)^2}{2(1) - 1}}{(2(1) - 1)^2} = \frac{0 - \frac{2(1)}{1}}{1^2} = -2$.

If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}$,then the value of $\frac{d^2 y}{d x^2}$ at the point $(\pi, 1)$ is

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