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

The derivative of (4-2x(-x^3-2x^2+1)) is 8x^3-16x-2

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

The first derivative of (4-2x(-x^3-2x^2+1)) is 8x^3-16x-2

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\frac{d}{d x} ((4 - 2 x) (- x^{3} - 2 x^{2} + 1))

8 x^{3} - 16 x - 2

Solution steps

\frac{d}{d x} ((4 - 2 x) (- x^{3} - 2 x^{2} + 1))

Apply the Product Rule:

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

f = 4 - 2 x, g = - x^{3} - 2 x^{2} + 1

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

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

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

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

Simplify

(- 2) (- x^{3} - 2 x^{2} + 1) + (- 3 x^{2} - 4 x) (4 - 2 x) : 8 x^{3} - 16 x - 2

= 8 x^{3} - 16 x - 2

\frac{\partial}{\partial x} (x e^{2 y} sin (5 z))

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

\frac{\partial}{\partial x} (x sin (y) cos (z))

y = cos (sin (x^{2}))

x \to 0 + lim (ln (x) \cdot 0)

What is the derivative of (4-2x(-x^3-2x^2+1)) ?
The derivative of (4-2x(-x^3-2x^2+1)) is 8x^3-16x-2
What is the first derivative of (4-2x(-x^3-2x^2+1)) ?
The first derivative of (4-2x(-x^3-2x^2+1)) is 8x^3-16x-2