Find dy dx in the following: ^ 2 x+ ^ 2 y=1 - Vidyadip

Question

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

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Answer

The given relationship is $\sin^{2}\text{x}+\cos^{2}\text{y}=1$
differenting this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\sin^{2}\text{x}+\cos^{2}\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin^{2}\text{x})+\frac{\text{d}}{\text{dx}}(\cos^{2}\text{y})=0$
$\Rightarrow 2\sin\text{x}.\frac{\text{d}}{\text{dx}}(\sin\text{x})+2\cos\text{y}.\frac{\text{d}}{\text{dx}}(\cos\text{y})=0$
$\Rightarrow2\sin\text{x}\cos\text{x}+2\cos\text{y}(-\sin\text{y}).\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\sin2\text{x}-\sin2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\sin2\text{x}}{\sin2\text{y}}$

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