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

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

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

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

2 x + 2 y

Solution steps

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

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

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

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

= 2 x + 2 y - 0

Simplify

= 2 x + 2 y

\int 1 + 100 x^{2} d x

\frac{1}{2} y^{^{'}} + 2 y = 0

\int_{x}^{\infty} 2 e^{- x - y} d y

\int (8 x - 7) ln (x) d x

y = 4 x^{2} - 13 x - \frac{9}{x ^{4}}

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