What is (\partial)/(\partial x)(12x+6y-x^2+xy-y^2) ?

The answer to (\partial)/(\partial x)(12x+6y-x^2+xy-y^2) is y+12-2x

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(\partial)/(\partial x)(12x+6y-x^2+xy-y^2)

\frac{\partial}{\partial x} (12 x + 6 y - x^{2} + x y - y^{2})

y + 12 - 2 x

Solution steps

\frac{\partial}{\partial x} (12 x + 6 y - x^{2} + x y - y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (12 x) + \frac{\partial}{\partial x} (6 y) - \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (x y) - \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (12 x) = 12

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

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

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

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

= 12 + 0 - 2 x + y - 0

Simplify

= y + 12 - 2 x

7 x - 6

\frac{\partial}{\partial x} (\frac{x y}{z})

\frac{d}{d x} (7 + x sec (x))

\int x e^{- x} d x

\frac{d}{d x} (64 e^{4 x})

What is (\partial)/(\partial x)(12x+6y-x^2+xy-y^2) ?
The answer to (\partial)/(\partial x)(12x+6y-x^2+xy-y^2) is y+12-2x