What is the derivative of (x^3/3+8/3-x^2+4) ?

The derivative of (x^3/3+8/3-x^2+4) is x^2-2x

What is the first derivative of (x^3/3+8/3-x^2+4) ?

The first derivative of (x^3/3+8/3-x^2+4) is x^2-2x

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\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{8}{3} - x^{2} + 4)

x^{2} - 2 x

Solution steps

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{8}{3} - x^{2} + 4)

Simplify

\frac{x ^{3}}{3} + \frac{8}{3} - x^{2} + 4 : \frac{x ^{3} + 8}{3} - x^{2} + 4

= \frac{d}{d x} (\frac{x ^{3} + 8}{3} - x^{2} + 4)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3} + 8}{3}) - \frac{d}{d x} (x^{2}) + \frac{d}{d x} (4)

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

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

\frac{d}{d x} (4) = 0

= x^{2} - 2 x + 0

Simplify

= x^{2} - 2 x

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'}} = 6 x

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

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

y^{^{'}} = x e^{x - y}

What is the derivative of (x^3/3+8/3-x^2+4) ?
The derivative of (x^3/3+8/3-x^2+4) is x^2-2x
What is the first derivative of (x^3/3+8/3-x^2+4) ?
The first derivative of (x^3/3+8/3-x^2+4) is x^2-2x