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

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

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

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

y - 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

= 0 - 2 x + y - 0

Simplify

= y - 2 x

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2 e^{5 x} cos (6 x)

\frac{1}{2} (t - 2)^{2}

\frac{\partial}{\partial x} (4 y + x e^{\frac{x}{y}})

y^{^{''}} - 4 y^{^{'}} + 5 y = 3 e^{2 t} sin (t)

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