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

(\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}

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

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

\frac{d}{d A} (0.0155 A^{3} - 0.372 A^{2} + 3.95 A + 1.21)

\int_{- 3}^{- 1} \frac{1}{x} d x

d er i v a t i v e x - \frac{1}{x}

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