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

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

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(\partial)/(\partial x)(x^2+2xy+2y^2-4x+4y)

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 4 x + 4 y)

2 x + 2 y - 4

Solution steps

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 4 x + 4 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\frac{\partial}{\partial x} (4 y) = 0

= 2 x + 2 y + 0 - 4 + 0

Simplify

= 2 x + 2 y - 4

f (x) = ln (x), at x = 2

\frac{d}{d x} (x^{2} + 2 x + 3)

\frac{d}{d x} (x^{\frac{3}{2}} - 4)

\frac{d}{d x} (\frac{x}{x + 6})

\int_{3}^{6} 21 - \frac{x ^{2}}{2} d x

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