8. Find dy/dx  in the following:

 \sin ^2 x + \cos ^ 2 y = 1 

Answers (1)

Given function is
\sin ^2 x + \cos ^ 2 y = 1
We can rewrite it as
\cos ^ 2 y = 1-\sin^2x
Now, differentiation w.r.t. x is
\frac{d(\cos^2y)}{dx} = \frac{d(1-\sin^2x)}{dx}
2\cos y (-\sin y)\frac{dy}{dx} = -2\sin x \cos x\\ \frac{dy}{dx} = \frac{2\sin x\cos x}{2\sin y \cos y} = \frac{\sin 2x }{\sin 2y} \ \ \ \ \ \ (\because2\sin a \cos a = \sin 2a)
Therefore, the answer is  \frac{\sin 2x}{\sin 2y }

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