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

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

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

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

3 x^{2} - 3 y

Solution steps

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

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

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

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

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

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

Simplify

= 3 x^{2} - 3 y

f (x) = 7 e^{x} cos (x), at x = 0

x \to 0 lim ((\frac{x - 4}{x + 1})^{x - 2})

\frac{d}{d x} (\frac{x x}{x ^{2}})

y^{^{'}} + 3 y = e^{- 4 t}, y (0) = 5

\frac{d}{d x} (\frac{9}{e ^{x} - e ^{- x}})

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