What is (\partial)/(\partial x)((2x-x^2)(2y-y^2)) ?

The answer to (\partial)/(\partial x)((2x-x^2)(2y-y^2)) is (2y-y^2)(2-2x)

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(\partial)/(\partial x)((2x-x^2)(2y-y^2))

\frac{\partial}{\partial x} ((2 x - x^{2}) (2 y - y^{2}))

(2 y - y^{2}) (2 - 2 x)

Solution steps

\frac{\partial}{\partial x} ((2 x - x^{2}) (2 y - y^{2}))

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= (2 y - y^{2}) \frac{\partial}{\partial x} (2 x - x^{2})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= (2 y - y^{2}) (\frac{\partial}{\partial x} (2 x) - \frac{\partial}{\partial x} (x^{2}))

\frac{\partial}{\partial x} (2 x) = 2

\frac{\partial}{\partial x} (x^{2}) = 2 x

= (2 y - y^{2}) (2 - 2 x)

\frac{\partial}{\partial x} (x^{yy - xx})

x \to 1 lim (\frac{2 x}{x ^{3} + 1})

\frac{d}{d x} ((\frac{- 2 x}{1 + x ^{2}})^{2})

2 x y y^{^{'}} = 1 + y^{2}

x \to 0 lim (- 5 (x - 3) ln (x - 3))

What is (\partial)/(\partial x)((2x-x^2)(2y-y^2)) ?
The answer to (\partial)/(\partial x)((2x-x^2)(2y-y^2)) is (2y-y^2)(2-2x)