What is the derivative of (2x^3/3-8x-5) ?

The derivative of (2x^3/3-8x-5) is 2x^2-8

What is the first derivative of (2x^3/3-8x-5) ?

The first derivative of (2x^3/3-8x-5) is 2x^2-8

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\frac{d}{d x} (\frac{2 x ^{3}}{3} - 8 x - 5)

2 x^{2} - 8

Solution steps

\frac{d}{d x} (\frac{2 x ^{3}}{3} - 8 x - 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{2 x ^{3}}{3}) - \frac{d}{d x} (8 x) - \frac{d}{d x} (5)

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

\frac{d}{d x} (8 x) = 8

\frac{d}{d x} (5) = 0

= 2 x^{2} - 8 - 0

Simplify

= 2 x^{2} - 8

y = tan (7 x), y = 2 sin (7 x)

x \frac{d y}{d x} + y = y^{2}

x^{2} + x

\int_{- 2}^{2} (x^{3}) d x

\frac{d}{d x} (\frac{54}{x})

What is the derivative of (2x^3/3-8x-5) ?
The derivative of (2x^3/3-8x-5) is 2x^2-8
What is the first derivative of (2x^3/3-8x-5) ?
The first derivative of (2x^3/3-8x-5) is 2x^2-8