derivative of |x^3+8x^2+5x-50|

Lösung

\frac{d}{d x} (x^{3} + 8 x^{2} + 5 x - 50)

\frac{( x ^{3} + 8 x ^{2} + 5 x - 50 ) ( 3 x ^{2} + 16 x + 5 )}{∣ x ^{3} + 8 x ^{2} + 5 x - 50 ∣}

Schritte zur Lösung

\frac{d}{d x} (x^{3} + 8 x^{2} + 5 x - 50)

Wende die Kettenregel an:

\frac{x ^{3} + 8 x ^{2} + 5 x - 50}{∣ ( x ^{3} + 8 x ^{2} + 5 x - 50 ) ∣} \frac{d}{d x} (x^{3} + 8 x^{2} + 5 x - 50)

= \frac{x ^{3} + 8 x ^{2} + 5 x - 50}{∣ ( x ^{3} + 8 x ^{2} + 5 x - 50 ) ∣} \frac{d}{d x} (x^{3} + 8 x^{2} + 5 x - 50)

\frac{d}{d x} (x^{3} + 8 x^{2} + 5 x - 50) = 3 x^{2} + 16 x + 5

= \frac{x ^{3} + 8 x ^{2} + 5 x - 50}{∣ ( x ^{3} + 8 x ^{2} + 5 x - 50 ) ∣} (3 x^{2} + 16 x + 5)

Multipliziere Brüche:

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

= \frac{( x ^{3} + 8 x ^{2} + 5 x - 50 ) ( 3 x ^{2} + 16 x + 5 )}{∣ x ^{3} + 8 x ^{2} + 5 x - 50 ∣}

\int \frac{1}{2 + 8 x} d x

\int sec (13 w) tan (13 w) d w

\int_{0}^{\frac{π}{4}} tan^{3} (x) d x

l a pl a ce t r an s f or m sin (4 t) cos (3 t)

\int 10 x - 3 d x