derivative of |x^3+2x^2-4x-8|

Lösung

\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 ∣}

Schritte zur Lösung

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

Wende die Kettenregel an:

\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)

Multipliziere Brüche:

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 ∣}