- #1

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^{2}y

^{2}+ (y+1)e

^{-x}=2 + x

Defines y as a differentiable function of x at point (x, y) = (0,1)

Find y′:

My attempt:

∂y/∂x =2xy

^{3}+ (-y-1)e

^{-x}=1

∂y/∂y = 3x

^{2}y

^{2}- e

^{-x}=0

Plugging in for x and y ⇒

∂y/∂x = -3

∂y/∂x = -1

For some reason I think y′ is defined as

(∂y/∂x) /(∂y/∂y) = 3

At leas this give the right answer (from solutions), but I am uncertain so I hope someone can correct me where I am wrong.