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

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

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

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

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derivative of

- 2 x^{2} (x + 4)

- 2 (3 x^{2} + 8 x)

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

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

f = x^{2}, g = x + 4

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

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

\frac{d}{d x} (x + 4) = 1

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

Simplify

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

= - 2 (3 x^{2} + 8 x)

\frac{\partial}{\partial x} (4 (x y)^{2})

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) (x + 7 x^{3}))

\int_{- 1}^{1} x^{2} - 2 x + 3 d x

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

\int_{0}^{\frac{π}{4}} \frac{1}{cos ( x )} d x

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