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

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

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

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

4 y^{3} - 4 x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= 4 y^{3} - 4 x

x \to 0 + lim ((sin (x))^{2 x})

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\frac{d}{d x} (x^{2} + y^{2} - 25)

\frac{\partial}{\partial x} (3 e^{5 x})

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