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

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

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

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

2 y - 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

= 2 y + 0 - 2 x - 0

Simplify

= 2 y - 2 x

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

\int \frac{2 x}{x} d x

\int \frac{sin ( 2 x )}{17 + cos ^{2} ( x )} d x

\frac{\partial}{\partial x} (4 y cos (9 x))

- x^{2} + 3, 2 x + 3

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