What is the derivative of 1/3 x^3+1/2 x^2-6x+8 ?

The derivative of 1/3 x^3+1/2 x^2-6x+8 is x^2+x-6

What is the first derivative of 1/3 x^3+1/2 x^2-6x+8 ?

The first derivative of 1/3 x^3+1/2 x^2-6x+8 is x^2+x-6

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\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8)

x^{2} + x - 6

Solution steps

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8)

Simplify

\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8 : \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 8

= \frac{d}{d x} (\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 8)

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} (8)

\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

\frac{d}{d x} (8) = 0

= x^{2} + x - 6 + 0

Simplify

= x^{2} + x - 6

\int x^{2} x^{3} + 35 d x

\frac{d}{d x} (1 + 1 - x)

y^{^{'}} - 5 y = 2 e^{6 t}

y = 2 x, y = x^{2} - 4 x

\frac{\partial}{\partial x} (- 7 x^{9})

What is the derivative of 1/3 x^3+1/2 x^2-6x+8 ?
The derivative of 1/3 x^3+1/2 x^2-6x+8 is x^2+x-6
What is the first derivative of 1/3 x^3+1/2 x^2-6x+8 ?
The first derivative of 1/3 x^3+1/2 x^2-6x+8 is x^2+x-6