What is the derivative of (x^3/3-4x) ?

The derivative of (x^3/3-4x) is x^2-4

What is the first derivative of (x^3/3-4x) ?

The first derivative of (x^3/3-4x) is x^2-4

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\frac{d}{d x} (\frac{x ^{3}}{3} - 4 x)

x^{2} - 4

Solution steps

\frac{d}{d x} (\frac{x ^{3}}{3} - 4 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{3}) - \frac{d}{d x} (4 x)

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

\frac{d}{d x} (4 x) = 4

= x^{2} - 4

f (x) = \frac{2 x - 3}{x ^{2} + 2 x}

2 x^{3} - e^{x y} = x \cdot y^{4}

x \to 0 + lim (x^{9 x})

\frac{\partial}{\partial x} (5 x^{2})

\int \frac{x}{6 - x ^{2}} d x

What is the derivative of (x^3/3-4x) ?
The derivative of (x^3/3-4x) is x^2-4
What is the first derivative of (x^3/3-4x) ?
The first derivative of (x^3/3-4x) is x^2-4