What is the derivative of f(x)= 8/3 x^3-16x^2+24x+4 ?

The derivative of f(x)= 8/3 x^3-16x^2+24x+4 is 8x^2-32x+24

What is the first derivative of f(x)= 8/3 x^3-16x^2+24x+4 ?

The first derivative of f(x)= 8/3 x^3-16x^2+24x+4 is 8x^2-32x+24

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derivative of

f (x) = \frac{8}{3} x^{3} - 16 x^{2} + 24 x + 4

8 x^{2} - 32 x + 24

Solution steps

\frac{d}{d x} (\frac{8}{3} x^{3} - 16 x^{2} + 24 x + 4)

Simplify

\frac{8}{3} x^{3} - 16 x^{2} + 24 x + 4 : \frac{8 x ^{3}}{3} - 16 x^{2} + 24 x + 4

= \frac{d}{d x} (\frac{8 x ^{3}}{3} - 16 x^{2} + 24 x + 4)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{8 x ^{3}}{3}) - \frac{d}{d x} (16 x^{2}) + \frac{d}{d x} (24 x) + \frac{d}{d x} (4)

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

\frac{d}{d x} (16 x^{2}) = 32 x

\frac{d}{d x} (24 x) = 24

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

= 8 x^{2} - 32 x + 24 + 0

Simplify

= 8 x^{2} - 32 x + 24

\frac{\partial}{\partial x} (3 + x)

y = e^{x}, at x = ln (2)

x \to 1 lim (5 - x)

f (x) = 3 cos^{2} (x)

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

What is the derivative of f(x)= 8/3 x^3-16x^2+24x+4 ?
The derivative of f(x)= 8/3 x^3-16x^2+24x+4 is 8x^2-32x+24
What is the first derivative of f(x)= 8/3 x^3-16x^2+24x+4 ?
The first derivative of f(x)= 8/3 x^3-16x^2+24x+4 is 8x^2-32x+24