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Question
Find `bb(dy/dx)` in the following:
sin2 y + cos xy = k
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Solution
sin2 y + cos xy = k
Differentiating both sides with respect to x,
`d/dx (sin^2 y) + d/dx (cos xy) = d/dx (k)`
⇒ `2 sin y cos y dy/dx + (- sin xy) d/dx (xy) = 0`
⇒ `2 sin y cos y dy/dx - sin xy [x dy/dx + y d/dx (x)] = 0`
⇒ `2 sin y cos y dy/dx - x sin xy dy/dx - y sin xy = 0`
⇒ `sin 2y - x sin xy dy/dx - y sin xy = 0`
⇒ `dy/dx (sin 2y - x sin xy) = y sin xy`
∴ `dy/dx = (y sin xy)/((sin 2y - x sin xy))`
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