What is the derivative of f(x)=(-2x^3+x^2-2x-1)x^3 ?

The derivative of f(x)=(-2x^3+x^2-2x-1)x^3 is -12x^5+5x^4-8x^3-3x^2

What is the first derivative of f(x)=(-2x^3+x^2-2x-1)x^3 ?

The first derivative of f(x)=(-2x^3+x^2-2x-1)x^3 is -12x^5+5x^4-8x^3-3x^2

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derivative of

f (x) = (- 2 x^{3} + x^{2} - 2 x - 1) x^{3}

- 12 x^{5} + 5 x^{4} - 8 x^{3} - 3 x^{2}

Solution steps

\frac{d}{d x} ((- 2 x^{3} + x^{2} - 2 x - 1) x^{3})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = - 2 x^{3} + x^{2} - 2 x - 1, g = x^{3}

= \frac{d}{d x} (- 2 x^{3} + x^{2} - 2 x - 1) x^{3} + \frac{d}{d x} (x^{3}) (- 2 x^{3} + x^{2} - 2 x - 1)

\frac{d}{d x} (- 2 x^{3} + x^{2} - 2 x - 1) = - 6 x^{2} + 2 x - 2

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

= (- 6 x^{2} + 2 x - 2) x^{3} + 3 x^{2} (- 2 x^{3} + x^{2} - 2 x - 1)

Simplify

(- 6 x^{2} + 2 x - 2) x^{3} + 3 x^{2} (- 2 x^{3} + x^{2} - 2 x - 1) : - 12 x^{5} + 5 x^{4} - 8 x^{3} - 3 x^{2}

= - 12 x^{5} + 5 x^{4} - 8 x^{3} - 3 x^{2}

y^{^{''}} + 64 y = sin (8 t)

x \to 0 lim (x^{2} + 4)

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

d y = x^{2} sin (8 x) d x

\frac{\partial}{\partial y} (x^{2} y^{3} sin (x - 2 y))

What is the derivative of f(x)=(-2x^3+x^2-2x-1)x^3 ?
The derivative of f(x)=(-2x^3+x^2-2x-1)x^3 is -12x^5+5x^4-8x^3-3x^2
What is the first derivative of f(x)=(-2x^3+x^2-2x-1)x^3 ?
The first derivative of f(x)=(-2x^3+x^2-2x-1)x^3 is -12x^5+5x^4-8x^3-3x^2