If y cos x + x cos y = pi,then y''(0) is | vedclass

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(B) Given the equation: $y \cos x + x \cos y = \pi$. At $x = 0$,we have $y \cos(0) + 0 \cos y = \pi$,which implies $y = \pi$. Differentiating both sid

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$\pi$

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(B) Given the equation: $y \cos x + x \cos y = \pi$. At $x = 0$,we have $y \cos(0) + 0 \cos y = \pi$,which implies $y = \pi$. Differentiating both sides with respect to $x$: $-y \sin x + y' \cos x + \cos y - x \sin y \cdot y' = 0$. At $x = 0$ and $y = \pi$: $-(\pi) \sin(0) + y'(0) \cos(0) + \cos(\pi) - 0 \sin(\pi) \cdot y'(0) = 0$. $0 + y'(0) \cdot 1 - 1 - 0 = 0 \implies y'(0) = 1$. Now,differentiate the first derivative expression again: $-y' \sin x - y \cos x + y'' \cos x - y' \sin x - \sin y \cdot y' - x(\cos y \cdot (y')^2 + \sin y \cdot y'') - \sin y \cdot y' = 0$. Substitute $x = 0, y = \pi, y' = 1$: $-1 \cdot 0 - \pi \cdot 1 + y''(0) \cdot 1 - 1 \cdot 0 - 0 - 0 - 0 = 0$. $-\pi + y''(0) = 0 \implies y''(0) = \pi$.

If $y \cos x + x \cos y = \pi$,then $y''(0)$ is

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