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

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

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

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

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

Solution steps

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

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

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

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

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

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

\int_{2.5}^{0.5} (23.75 - 7.5 v) d v

\frac{\partial}{\partial y} (8 x - (x + y)^{3})

\frac{d}{d x} (\frac{1}{x ^{31}})

(\frac{2}{x} - 1)^{- 1}

x \to 0 lim (ln (\frac{x}{3}) + x^{2} - 8)

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