What is (\partial)/(\partial x)((5x+6y)-(x^2-4xy+8y^2+7x-14y-6)) ?

The answer to (\partial)/(\partial x)((5x+6y)-(x^2-4xy+8y^2+7x-14y-6)) is 4y-2x-2

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(\partial)/(\partial x)((5x+6y)-(x^2-4xy+8y^2+7x-14y-6))

\frac{\partial}{\partial x} ((5 x + 6 y) - (x^{2} - 4 x y + 8 y^{2} + 7 x - 14 y - 6))

4 y - 2 x - 2

Solution steps

\frac{\partial}{\partial x} ((5 x + 6 y) - (x^{2} - 4 x y + 8 y^{2} + 7 x - 14 y - 6))

Simplify

(5 x + 6 y) - (x^{2} - 4 x y + 8 y^{2} + 7 x - 14 y - 6) : - x^{2} + 4 x y - 2 x - 8 y^{2} + 20 y + 6

= \frac{\partial}{\partial x} (- x^{2} + 4 x y - 2 x - 8 y^{2} + 20 y + 6)

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} (4 x y) - \frac{\partial}{\partial x} (2 x) - \frac{\partial}{\partial x} (8 y^{2}) + \frac{\partial}{\partial x} (20 y) + \frac{\partial}{\partial x} (6)

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

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

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

\frac{\partial}{\partial x} (8 y^{2}) = 0

\frac{\partial}{\partial x} (20 y) = 0

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

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

Simplify

= 4 y - 2 x - 2

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What is (\partial)/(\partial x)((5x+6y)-(x^2-4xy+8y^2+7x-14y-6)) ?
The answer to (\partial)/(\partial x)((5x+6y)-(x^2-4xy+8y^2+7x-14y-6)) is 4y-2x-2