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

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

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

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

6 y^{2} x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= - 0 + 6 y^{2} x - 0

Simplify

= 6 y^{2} x

y^{^{'}} = 24 x^{2} e^{- y}

\int 4 - 3 (1 - x^{2})^{- 1} d x

\frac{d}{d x} y - y = 4

f (x) = \frac{- 2 x}{x ^{2} + 1}

\int \frac{5 - 50 x}{4 - 25 x ^{2}} d x

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