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प्रश्न
Find `bb(dy/dx)` in the following:
sin2 x + cos2 y = 1
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उत्तर
sin2 x + cos2 y = 1
Differentiating both sides with respect to x,
`d/dx sin^2 x + d/dx cos^2 y = d/dx (1)`
⇒ `2 sin x d/dx sin x + 2 cos y d/dx cos y = 0`
⇒ `2 sin x cos x + 2 cos y (- sin y) dy/dx = 0`
⇒ `2 sin x cos x - 2 cos y sin y dy/dx = 0`
⇒ `sin 2x - sin 2y dy/dx = 0`
∴ `dy/dx = (sin 2x) /(sin 2y)`
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