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

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

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

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

- 6 x y

Solution steps

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

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

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

= 0 - 6 x y

Simplify

= - 6 x y

1 - (1 - 0.01 t)^{3}

\int \frac{ln ( x )}{x ^{8}} d x

\int_{0}^{1} \frac{33}{4 y - 1} d y

\int_{0}^{\infty} 2 d x

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

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