- (C) Differentiating implicitly yields 4x − 3y^2 y ′ = 0. So The linear approximation for
the true value of y when x changes from 3 to 3.04 is
Since it is given that, when x = 3, y = 2, the approximate value of y is
or
Since it is given that, when x = 3, y = 2, the approximate value of y is
or