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

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

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

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

2 y - 2 x

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 y - 2 x

\int csc^{4} (3 x) d x

\int \frac{x}{1 - x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x ^{2} - 6})

\frac{d y}{d x} + 3 y = e^{4 x}

\frac{\partial}{\partial x} (2 k^{3} s^{2})

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