What is the derivative of |x^3+2x^2-4x-8| ?

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (x^{3} + 2 x^{2} - 4 x - 8)

\frac{( x ^{3} + 2 x ^{2} - 4 x - 8 ) ( 3 x ^{2} + 4 x - 4 )}{∣ x ^{3} + 2 x ^{2} - 4 x - 8 ∣}

Solution steps

\frac{d}{d x} (x^{3} + 2 x^{2} - 4 x - 8)

Apply the chain rule:

\frac{x ^{3} + 2 x ^{2} - 4 x - 8}{∣ ( x ^{3} + 2 x ^{2} - 4 x - 8 ) ∣} \frac{d}{d x} (x^{3} + 2 x^{2} - 4 x - 8)

= \frac{x ^{3} + 2 x ^{2} - 4 x - 8}{∣ ( x ^{3} + 2 x ^{2} - 4 x - 8 ) ∣} \frac{d}{d x} (x^{3} + 2 x^{2} - 4 x - 8)

\frac{d}{d x} (x^{3} + 2 x^{2} - 4 x - 8) = 3 x^{2} + 4 x - 4

= \frac{x ^{3} + 2 x ^{2} - 4 x - 8}{∣ ( x ^{3} + 2 x ^{2} - 4 x - 8 ) ∣} (3 x^{2} + 4 x - 4)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( x ^{3} + 2 x ^{2} - 4 x - 8 ) ( 3 x ^{2} + 4 x - 4 )}{∣ x ^{3} + 2 x ^{2} - 4 x - 8 ∣}

x \to 4 lim (3 e^{1.7 x} + 1)

y^{^{''}} - 4 y^{^{'}} + 4 y = 8 x^{2} + e^{2 x}

y = 3 x^{3} + 45 x^{2} + 81 x

\int \frac{1}{x x + 4} d x

y = sin^{2} (x), y = sin^{3} (x)

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