What is the derivative of 64x-(x^3/3) ?

The derivative of 64x-(x^3/3) is 64-x^2

What is the first derivative of 64x-(x^3/3) ?

The first derivative of 64x-(x^3/3) is 64-x^2

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\frac{d}{d x} (64 x - \frac{x ^{3}}{3})

64 - x^{2}

Solution steps

\frac{d}{d x} (64 x - \frac{x ^{3}}{3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (64 x) - \frac{d}{d x} (\frac{x ^{3}}{3})

\frac{d}{d x} (64 x) = 64

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

= 64 - x^{2}

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

x \to 1 lim ((x - 2)^{2})

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

\frac{\partial}{\partial x} (ln (x^{2} + 2 x))

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

What is the derivative of 64x-(x^3/3) ?
The derivative of 64x-(x^3/3) is 64-x^2
What is the first derivative of 64x-(x^3/3) ?
The first derivative of 64x-(x^3/3) is 64-x^2