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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)`
Posted by justaguide on May 20, 2012 at 6:00 PM (Answer #1)
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