Find the derivative y' given that `xe^y-3ysinx=1`

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The derivative y' has to be determined given that `xe^y-3ysinx=1` . Using implicit differentiation

`e^y + x*e^y*y' - 3y'*sin x - 3ycos x = 0`

=> `y'(x*e^y - 3sin x) = 3y*cos x - e^y`

=> ` y' = (3y*cos x - e^y)/(x*e^y - 3sin x)`

The derivative `y' = (3y*cos x - e^y)/(x*e^y - 3sin x)`

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