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

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

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

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

3 x^{2} - 3 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= 3 x^{2} - 3 y

x \to - 2 lim (\frac{1}{∣ x + 2 ∣} + x^{2})

y^{^{'}} = 2.5 - \frac{y}{2000}

\frac{d y}{d t} + \frac{1}{2} y = 17 sin (2 t), y (0) = - 1

\int x^{- 2} + x^{\frac{1}{5}} d x

\frac{d}{d x} (arctan (s x))

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