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Find $ \frac {dy}{dx} $ in the following: $ \sin^2x +\cos^2y = 1 $

$\begin{array}{1 1}\large \frac{\sin\;2x}{\sin\;2y} \\ -\large \frac{\sin\;2x}{\sin\;2y} \\ \large \frac{\cos\;2x}{\cos\;2y} \\ -\large \frac{\cos\;2x}{\cos\;2y} \end{array} $

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