What is (\partial)/(\partial x)(x^2+2xy+2y^2-6x+6y) ?

The answer to (\partial)/(\partial x)(x^2+2xy+2y^2-6x+6y) is 2x+2y-6

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(\partial)/(\partial x)(x^2+2xy+2y^2-6x+6y)

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 6 x + 6 y)

2 x + 2 y - 6

Solution steps

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 6 x + 6 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} (6 x) + \frac{\partial}{\partial x} (6 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} (6 x) = 6

\frac{\partial}{\partial x} (6 y) = 0

= 2 x + 2 y + 0 - 6 + 0

Simplify

= 2 x + 2 y - 6

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

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

g^{'} (x)

\int - \frac{6}{x} d x

x \to 0 - lim (arctan (\frac{1}{x}))

What is (\partial)/(\partial x)(x^2+2xy+2y^2-6x+6y) ?
The answer to (\partial)/(\partial x)(x^2+2xy+2y^2-6x+6y) is 2x+2y-6