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

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

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

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

2 x + 2 y - 8

Solution steps

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

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

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

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

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

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

= 2 x + 2 y - 0 - 8

Simplify

= 2 x + 2 y - 8

f (t) = - 4 t^{2} + 16 t + 9

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\int 15 sin^{2} (x) cos^{3} (x) d x

\frac{d}{d x} (\frac{7}{2 x})

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