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

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

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

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

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

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

Solution steps

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

Simplify

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

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

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

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

\int (sin (x) + 2 cos (x)) d x

\int \frac{x e ^{x}}{1 + e ^{x}} d x

(\frac{x}{1 - x})^{^{'}}

\int cos (7 x) d x

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0

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