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

The answer to (\partial)/(\partial y)(x^4+y^4-4xy+2) is 4y^3-4x

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

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

4 y^{3} - 4 x

Solution steps

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

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} (2)

\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} (2) = 0

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

Simplify

= 4 y^{3} - 4 x

x + \frac{5}{x}

\int_{0}^{1} \frac{10}{4 y - 1} d y

x \to - 1 lim (ln (x))

\frac{\partial}{\partial x} (4 x^{0.5} y^{0.5})

y = sin^{2} (x^{3})

What is (\partial)/(\partial y)(x^4+y^4-4xy+2) ?
The answer to (\partial)/(\partial y)(x^4+y^4-4xy+2) is 4y^3-4x