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`

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