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

The derivative of (4x^3/3-x) is 4x^2-1

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

The first derivative of (4x^3/3-x) is 4x^2-1

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

4 x^{2} - 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

= 4 x^{2} - 1

y = (x - 2) (x + 1) (x + 2)

\int a^{- 3 x} d x

\int \frac{e ^{3 x}}{( 4 + e ^{2 x} ) ^{2}} d x

f (x) = \frac{8}{x + 11}, at x = 5

\int_{0}^{1} 3 x d x

What is the derivative of (4x^3/3-x) ?
The derivative of (4x^3/3-x) is 4x^2-1
What is the first derivative of (4x^3/3-x) ?
The first derivative of (4x^3/3-x) is 4x^2-1