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

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

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(\partial)/(\partial y)(x^3+3x^2y-y^4)

\frac{\partial}{\partial y} (x^{3} + 3 x^{2} y - y^{4})

3 x^{2} - 4 y^{3}

Solution steps

\frac{\partial}{\partial y} (x^{3} + 3 x^{2} y - y^{4})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{3}) + \frac{\partial}{\partial y} (3 x^{2} y) - \frac{\partial}{\partial y} (y^{4})

\frac{\partial}{\partial y} (x^{3}) = 0

\frac{\partial}{\partial y} (3 x^{2} y) = 3 x^{2}

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

= 0 + 3 x^{2} - 4 y^{3}

Simplify

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

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