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Question 7.1.3: The equation x²y³ + (y + 1)e^−x = x + 2 defines y as a diffe......

The equation x^{2}y^{3} + (y + 1)e^{−x} = x + 2 defines y as a differentiable function of x in a neighbourhood of (x, y) = (0, 1). Compute y^{\prime} at this point.

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Implicit differentiation w.r.t. x gives

2x y^{3}+x^{2}3y^{2}y^{\prime}+y^{\prime}e^{-x}+(y+1)(-e^{-x})=1

Inserting x = 0 and y = 1 yields y + 2(−1) = 1, implying that y^{\prime} = 3.

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