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

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

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

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

4 x - 2 x y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 4 x + 0 - 2 y x

Simplify

= 4 x - 2 x y

\frac{\partial}{\partial x} (sin (\frac{1}{x ^{2} + y ^{2}}))

\frac{\partial}{\partial x} (y e^{x})

\frac{d}{d x} (3 sin (x) - 2 cos (x))

x \to 3 lim ((4 x + 5) (6 x - 7))

\int x 1 + y^{3} d y

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