What is (\partial)/(\partial y)(x^4-x^2-2xy+y^2+1) ?

The answer to (\partial)/(\partial y)(x^4-x^2-2xy+y^2+1) is 2y-2x

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(\partial)/(\partial y)(x^4-x^2-2xy+y^2+1)

\frac{\partial}{\partial y} (x^{4} - x^{2} - 2 x y + y^{2} + 1)

2 y - 2 x

Solution steps

\frac{\partial}{\partial y} (x^{4} - x^{2} - 2 x y + y^{2} + 1)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{4}) - \frac{\partial}{\partial y} (x^{2}) - \frac{\partial}{\partial y} (2 x y) + \frac{\partial}{\partial y} (y^{2}) + \frac{\partial}{\partial y} (1)

\frac{\partial}{\partial y} (x^{4}) = 0

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

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

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

\frac{\partial}{\partial y} (1) = 0

= 0 - 0 - 2 x + 2 y + 0

Simplify

= 2 y - 2 x

\frac{d}{d x} (e^{4 e^{x}})

\int \frac{1}{x x ^{4} + 1} d x

\int \frac{2 x}{16 x ^{4} - 9} d x

2 x e^{x} cos (x)

\frac{d}{d x} (ln (\frac{x ^{\frac{1}{2}}}{3}))

What is (\partial)/(\partial y)(x^4-x^2-2xy+y^2+1) ?
The answer to (\partial)/(\partial y)(x^4-x^2-2xy+y^2+1) is 2y-2x