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If $y \cos x+x \cos y=\pi$, then $y^{\prime \prime}(0)$ is
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$\pi$
$y \cos x+x \cos y=\pi$
Differentiate both sides with respect to $x$, we get
$-y \sin x+\cos x \cdot y^{\prime}+x(-\sin y) y^{\prime}+\cos y$
Again differentiate with respect to $x$
$\begin{aligned} &-y^{\prime \prime} \sin x-y \cos x+\cos x \cdot y^{\prime \prime}+\sin x \cdot y^{\prime}-\sin y \cdot y^{\prime} \\ &-x\left[\cos y \cdot\left(y^{\prime}\right)^2+\sin y \cdot y^{\prime \prime}\right]-\sin y \cdot y^{\prime}\end{aligned}$
Putting $x=0$, we get $-y+y^{\prime \prime}-2 \sin y y^{\prime}=0$
$y^{\prime \prime}=y+2 y^{\prime} \sin y$
Since at $x=0, y=\pi ;\left(y^{\prime \prime}\right)_0=\pi$.
Differentiate both sides with respect to $x$, we get
$-y \sin x+\cos x \cdot y^{\prime}+x(-\sin y) y^{\prime}+\cos y$
Again differentiate with respect to $x$
$\begin{aligned} &-y^{\prime \prime} \sin x-y \cos x+\cos x \cdot y^{\prime \prime}+\sin x \cdot y^{\prime}-\sin y \cdot y^{\prime} \\ &-x\left[\cos y \cdot\left(y^{\prime}\right)^2+\sin y \cdot y^{\prime \prime}\right]-\sin y \cdot y^{\prime}\end{aligned}$
Putting $x=0$, we get $-y+y^{\prime \prime}-2 \sin y y^{\prime}=0$
$y^{\prime \prime}=y+2 y^{\prime} \sin y$
Since at $x=0, y=\pi ;\left(y^{\prime \prime}\right)_0=\pi$.
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