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

The derivative of (-x^3+1(-2x^2-2x)) is 10x^4+8x^3-4x-2

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

The first derivative of (-x^3+1(-2x^2-2x)) is 10x^4+8x^3-4x-2

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

= (- 3 x^{2}) (- 2 x^{2} - 2 x) + (- 4 x - 2) (- x^{3} + 1)

Simplify

(- 3 x^{2}) (- 2 x^{2} - 2 x) + (- 4 x - 2) (- x^{3} + 1) : 10 x^{4} + 8 x^{3} - 4 x - 2

= 10 x^{4} + 8 x^{3} - 4 x - 2

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 16 t e^{5 t}

y^{^{''}} + 6 y^{^{'}} - 4 y = 0

ln (1 0^{x})

f (x) = e^{x^{2} + 2}

What is the derivative of (-x^3+1(-2x^2-2x)) ?
The derivative of (-x^3+1(-2x^2-2x)) is 10x^4+8x^3-4x-2
What is the first derivative of (-x^3+1(-2x^2-2x)) ?
The first derivative of (-x^3+1(-2x^2-2x)) is 10x^4+8x^3-4x-2