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

The answer to (\partial)/(\partial y)(x^2+2y^2-x^2y) is 4y-x^2

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

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

4 y - x^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} (2 y^{2}) = 4 y

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

= 0 + 4 y - x^{2}

Simplify

= 4 y - x^{2}

x \to \frac{7 π}{2} + lim (sec (x))

\frac{d}{d x} ((2 x - 9)^{4} (x^{2} + x + 1)^{5})

\frac{d}{d x} (3 ln (15 x))

x \to 3 lim (cos (\frac{π}{4} x))

x \to a lim (\frac{4 ^{2} + 4}{x + 5})

What is (\partial)/(\partial y)(x^2+2y^2-x^2y) ?
The answer to (\partial)/(\partial y)(x^2+2y^2-x^2y) is 4y-x^2