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If π/2x32sin2zdz+0ycostdt=0 then  dydx at (π/2,π) is

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detailed solution

Correct option is A

Differentiating w ·r· t· x, we get3−2sin2⁡x+dydxcos⁡y=0Putting x=π/2 and y=π we getdydx(π/2,π)=1.


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