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

The answer to (\partial)/(\partial y)(x^3-xy^2+3y^2) is 6y-2xy

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

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

6 y - 2 x y

Solution steps

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

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

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

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

\frac{\partial}{\partial y} (3 y^{2}) = 6 y

= 0 - 2 x y + 6 y

Simplify

= 6 y - 2 x y

What is (\partial)/(\partial y)(x^3-xy^2+3y^2) ?
The answer to (\partial)/(\partial y)(x^3-xy^2+3y^2) is 6y-2xy