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

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

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

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

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derivative of

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

4 x^{3} + 4 x^{2} - 8 x

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= 4 x^{3} + 4 x^{2} - 8 x

\int \frac{y + 1}{y ^{2} + 1} d y

f (x) = 5 x^{3} - 5 x, (- 1, 0)

g (x) = \frac{1 + 2 x}{3 - 4 x}

\int x (4 x + 5)^{3} d x

\frac{d y}{d x}, x y = 28

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