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

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

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

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

3 y^{2} - 6 x y

Solution steps

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

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} (y^{3})

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

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

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

= 0 - 6 x y + 3 y^{2}

Simplify

= 3 y^{2} - 6 x y

f (x) = \frac{2 t + 3}{t + 5}

\frac{d ^{2}}{d x ^{2}} (t e^{- t})

\int x - 2 x d x

y = - 2 x^{3}, (- 1, 2)

\int_{0}^{1} \int_{0}^{1} 4 x y e^{- 2 x - y} d y d x

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