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

The answer to (\partial)/(\partial y)(3x^2y-y^3+x-y) is 3x^2-3y^2-1

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

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

3 x^{2} - 3 y^{2} - 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

= 3 x^{2} - 3 y^{2} + 0 - 1

Simplify

= 3 x^{2} - 3 y^{2} - 1

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What is (\partial)/(\partial y)(3x^2y-y^3+x-y) ?
The answer to (\partial)/(\partial y)(3x^2y-y^3+x-y) is 3x^2-3y^2-1