What is the derivative of (x^3/3-(x^2)/2-6x) ?

The derivative of (x^3/3-(x^2)/2-6x) is x^2-x-6

What is the first derivative of (x^3/3-(x^2)/2-6x) ?

The first derivative of (x^3/3-(x^2)/2-6x) is x^2-x-6

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

x^{2} - x - 6

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (6 x) = 6

= x^{2} - x - 6

\frac{d}{d x} (e^{x} x^{5} + 10 e^{x} x^{4} + 20 e^{x} x^{3})

\int 9 sin^{2} (x) cos (x) d x

f (x) = 4 x + 5

\int (x^{3} + cos (2 x)) d x

f (θ) = 2 sin (θ)

What is the derivative of (x^3/3-(x^2)/2-6x) ?
The derivative of (x^3/3-(x^2)/2-6x) is x^2-x-6
What is the first derivative of (x^3/3-(x^2)/2-6x) ?
The first derivative of (x^3/3-(x^2)/2-6x) is x^2-x-6