Use the chain rule for implicit differentiation:

`"ln" y = 2y^2 - x ``(1)/(y) y' = 4 y y' -1``((1)/(y) - 4y) y' = -1``y' = (-1)/((1)/(y) - 4y)=(-y)/(1-4y^2)=(y)/(4y^2-1)`At (2,1), `y' = (1)/(4(2)^2-1)=(1)/(15)`

Use the chain rule for implicit differentiation:

`"ln" y = 2y^2 - x `

`(1)/(y) y' = 4 y y' -1`

`((1)/(y) - 4y) y' = -1`

`y' = (-1)/((1)/(y) - 4y)=(-y)/(1-4y^2)=(y)/(4y^2-1)`

At (2,1),

`y' = (1)/(4(2)^2-1)=(1)/(15)`