partial derivative of (x-y(1-xy))

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∂∂x ((x−y)(1−xy))=1−2xy+y²

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∂ ∂ x ((x−y)(1−xy))=1−2xy+y²

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∂ ∂ x ((x−y)(1−xy))

Treat y as a constant

Apply the Product Rule: (f·g)′=f ′·g+f·g′
f=x−y, g=1−xy
=∂ ∂ x (x−y)(1−xy)+∂ ∂ x (1−xy)(x−y)

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∂ ∂ x (x−y)=1
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∂ ∂ x (1−xy)=−y
=1·(1−xy)+(−y)(x−y)

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Simplify 1·(1−xy)+(−y)(x−y): 1−2xy+y²
=1−2xy+y²