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Question:

Find $\frac{d y}{d x}$ :

$\sin ^{2} x+\cos ^{2} y=1$

Solution:

The given relationship is $\sin ^{2} x+\cos ^{2} y=1$

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}\left(\sin ^{2} x+\cos ^{2} y\right)=\frac{d}{d x}(1)$

$\Rightarrow \frac{d}{d x}\left(\sin ^{2} x\right)+\frac{d}{d x}\left(\cos ^{2} y\right)=0$

$\Rightarrow 2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \cos y \cdot \frac{d}{d x}(\cos y)=0$

$\Rightarrow 2 \sin x \cos x+2 \cos y(-\sin y) \cdot \frac{d y}{d x}=0$

$\Rightarrow \sin 2 x-\sin 2 y \frac{d y}{d x}=0$

$\therefore \frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}$

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