The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].
In this problem we have a function in implicit form, that is, an expression of the form: f(x, y, c) = 0, where c is a constant. Then, we should apply implicit differentiation twice to determine the second derivative of the function:
Original expression
x · y - 1 = 2 · x + y²
First derivative
y + x · y' = 2 + 2 · y · y'
(x - 2 · y) · y' = 2 - y
y' = (2 - y) / (x - 2 · y)
Second derivative
y' + y' + x · y'' = 2 · (y')² + 2 · y · y''
2 · y' - 2 · (y')² = (2 · y - x) · y''
y'' = 2 · [y' - (y')²] / (2 · y - x)
y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]]
The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].
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