What is the second derivative of xy+y^2=4
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Differentiating the given function w.r.t. x we find:

yd/dx(x) +xdy/dx+ 2ydy/dx = 0

i.e. y * 1 +xdy/dx+ 2ydy/dx = 0

i.e. y  + (x+ 2y) dy/dx = 0

i.e. dy/dx = -y/(x+ 2y)

Differentiating again w.r.t. x we find:

d^2y/dx^2 = [(x+ 2y)dy/dx(-y) + (-y) d/dx(x+ 2y)]/(x+ 2y)^2

d^2y/dx^2 = [-(x+ 2y) + (-y) (1+ 2dy/dx)]/(x+ 2y)^2

d^2y/dx^2 = [-x- 2y -y (1+ 2dy/dx)]/(x+ 2y)^2

d^2y/dx^2 = [-x- 2y -y - 2ydy/dx)]/(x+ 2y)^2

d^2y/dx^2 = [-x- 3y  - 2ydy/dx)]/(x+ 2y)^2


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