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

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

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(\partial)/(\partial y)(x^3-xy+y^3)

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

3 y^{2} - x

Solution steps

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

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

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

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

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

= 0 - x + 3 y^{2}

Simplify

= 3 y^{2} - x

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