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

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

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

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

3 x^{2} - 3

Solution steps

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

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

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

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

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

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

Simplify

= 3 x^{2} - 3

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