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Question

Find dydx of sin2y+cosxy=k

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Solution

We have, sin2y+cosxy=k
Differentiating both sides with respect to x, we obtain
ddx(sin2y)+ddx(cosxy)=d(π)dx=0 .....(1)
Using chain rule, we obtain
ddx(sin2y)=2sinyddx(siny)=2sinycosydydx...... (2)
and
ddx(cosxy)=sinxyddx(xy)=sinxy[yddx(x)+xdydx]
=sinxy[y.1+xdydx]=ysinxyxsinxydydx.....(3)
From (1), (2) and (3), we obtain
2sinycosydydxysinxyxsinxydydx=0
(2sinycosyxsinxy)dydx=ysinxy
(sin2yxsinxy)dydx=ysinxy
dydx=ysinxysin2yxsinxy

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