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

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

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

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

2 y - 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 2 y - 2 x

Simplify

= 2 y - 2 x

x \to 16 lim (x + 2 x)

y = x \cdot x

y = 4 x - 3 x^{2} (2 - 4)

3 x y^{3} d x - (x + 2) d y = 0

\frac{d}{d x} (2 cos^{2} (2 x))

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