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

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

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

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

6 x y - 3

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (3 x) = 3

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

= 6 y x - 3 + 0

Simplify

= 6 x y - 3

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