What is the derivative of 2x^3-4x ?

The derivative of 2x^3-4x is 6x^2-4

What is the first derivative of 2x^3-4x ?

The first derivative of 2x^3-4x is 6x^2-4

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\frac{d}{d x} (2 x^{3} - 4 x)

6 x^{2} - 4

Solution steps

\frac{d}{d x} (2 x^{3} - 4 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{3}) - \frac{d}{d x} (4 x)

\frac{d}{d x} (2 x^{3}) = 6 x^{2}

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

= 6 x^{2} - 4

\frac{2}{3 x ^{4}}

n = 1 \sum \infty \frac{n}{5 + n ^{2}}

2 x - x^{2}, - 2 x

\int \frac{1 + 16 x}{25 - x ^{2}} d x

\frac{d}{d x} (\frac{2 x}{3 y})

What is the derivative of 2x^3-4x ?
The derivative of 2x^3-4x is 6x^2-4
What is the first derivative of 2x^3-4x ?
The first derivative of 2x^3-4x is 6x^2-4