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

The answer to (\partial)/(\partial y)(x^4+y^4-4xy+1) is 4y^3-4x

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

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

4 y^{3} - 4 x

Solution steps

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

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

\frac{\partial}{\partial y} (y^{4}) = 4 y^{3}

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

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

= 0 + 4 y^{3} - 4 x + 0

Simplify

= 4 y^{3} - 4 x

\int x^{3} + x d x

\frac{d y}{d x} = 6 x^{2} e^{- y}

\int (y^{3} + 1.8 y^{2} - 2.4 y) d y

x \to - 2 lim (\frac{3 x - 1}{x})

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

What is (\partial)/(\partial y)(x^4+y^4-4xy+1) ?
The answer to (\partial)/(\partial y)(x^4+y^4-4xy+1) is 4y^3-4x