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Find d²y/dx² for implicitly in terms of x and y
xy-1=2x+y²

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xero099

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)]].

What is the second derivative of an implicit equation?

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)]].

To learn more on implicite differentiation: https://brainly.com/question/11887805

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