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If $\sin ^2 x+2 \cos y+x y=0$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{y+\sin 2 x}{2 \sin y-x}$
$\begin{aligned} & \sin ^2 x+2 \cos y+x y=0 \\ & \Rightarrow 2 \sin x \cos x-2 \sin y \frac{d y}{d x}+y+x \frac{d y}{d x}=0\end{aligned}$
$\therefore \frac{d y}{d x}=\frac{y+\sin 2 x}{2 \sin y-x}$.
$\therefore \frac{d y}{d x}=\frac{y+\sin 2 x}{2 \sin y-x}$.
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