What is the derivative of (x^4/4+8/9 x^3-x^2+3x-8) ?

The derivative of (x^4/4+8/9 x^3-x^2+3x-8) is x^3+(8x^2)/3-2x+3

What is the first derivative of (x^4/4+8/9 x^3-x^2+3x-8) ?

The first derivative of (x^4/4+8/9 x^3-x^2+3x-8) is x^3+(8x^2)/3-2x+3

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\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{8}{9} x^{3} - x^{2} + 3 x - 8)

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

Solution steps

\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{8}{9} x^{3} - x^{2} + 3 x - 8)

Simplify

\frac{x ^{4}}{4} + \frac{8}{9} x^{3} - x^{2} + 3 x - 8 : \frac{x ^{4}}{4} + \frac{8 x ^{3}}{9} - x^{2} + 3 x - 8

= \frac{d}{d x} (\frac{x ^{4}}{4} + \frac{8 x ^{3}}{9} - x^{2} + 3 x - 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{4}}{4}) + \frac{d}{d x} (\frac{8 x ^{3}}{9}) - \frac{d}{d x} (x^{2}) + \frac{d}{d x} (3 x) - \frac{d}{d x} (8)

\frac{d}{d x} (\frac{x ^{4}}{4}) = x^{3}

\frac{d}{d x} (\frac{8 x ^{3}}{9}) = \frac{8 x ^{2}}{3}

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

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

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

= x^{3} + \frac{8 x ^{2}}{3} - 2 x + 3 - 0

Simplify

= x^{3} + \frac{8 x ^{2}}{3} - 2 x + 3

x \to - 8 lim (\frac{∣ x - 8 ∣}{x - 8})

x \to 0 lim (\frac{x ^{3}}{3})

\frac{d}{d x} (- 5 x^{4})

x \to 3 + lim (\frac{b x - 3 b}{x - 3})

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

What is the derivative of (x^4/4+8/9 x^3-x^2+3x-8) ?
The derivative of (x^4/4+8/9 x^3-x^2+3x-8) is x^3+(8x^2)/3-2x+3
What is the first derivative of (x^4/4+8/9 x^3-x^2+3x-8) ?
The first derivative of (x^4/4+8/9 x^3-x^2+3x-8) is x^3+(8x^2)/3-2x+3