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Given ln y+2y^2-x Find y' and evaluate y' at (2,1)
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`lny + 2y^2-x = 0`
Using implicit differentiation,
`1/y xx dy + 2y xx dy - dx = 0`
`1/y xx dy + 2y xx dy = dx`
`(1/y+2y)dy = dx`
`(dy)/(dx) = 1/(1/y+2y)`
`(dy)/(dx) = y/(1+2y^2)`
at `(2,1), (dy)/(dx) = 1/(1+2xx1)`
`(dy)/(dx) = 1/(1+2) = 1/3`
Therefore at `(2,1), y' = 1/3`
Posted by thilina-g on June 18, 2012 at 9:39 AM (Answer #1)
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