What is (\partial)/(\partial x)(x^2+2xy+2y^2-8x+8y) ?

The answer to (\partial)/(\partial x)(x^2+2xy+2y^2-8x+8y) is 2x+2y-8

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(\partial)/(\partial x)(x^2+2xy+2y^2-8x+8y)

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 8 x + 8 y)

2 x + 2 y - 8

Solution steps

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 8 x + 8 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} (8 x) + \frac{\partial}{\partial x} (8 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} (8 x) = 8

\frac{\partial}{\partial x} (8 y) = 0

= 2 x + 2 y + 0 - 8 + 0

Simplify

= 2 x + 2 y - 8

\int_{1}^{\infty} \frac{1}{x ^{2} + 3} d x

y = 3 1.6 x^{2} + 4.9

f (t) = \frac{1}{2} (9 t^{2} + t)^{- 3}

x \to \infty lim (2 1^{x})

\frac{\partial}{\partial y} (cos (12 x) sin (2 y))

What is (\partial)/(\partial x)(x^2+2xy+2y^2-8x+8y) ?
The answer to (\partial)/(\partial x)(x^2+2xy+2y^2-8x+8y) is 2x+2y-8