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

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

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

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

3 y^{2} - 3 x

Solution steps

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

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

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

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

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

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

Simplify

= 3 y^{2} - 3 x

f (x) = 4 - 3 x, g (x) = 2 x^{2} - x

\int_{- 3}^{- 2} x^{2} e^{x^{3} + 27} d x

x \to 2 lim (2 x - 9)

\frac{\partial}{\partial x} (- \frac{x y}{( 1 + x ) ^{3}})

y = x (1.1)

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