derivative of |2x+1|

Lösung

\frac{d}{d x} (∣ 2 x + 1 ∣)

\frac{2 ( 2 x + 1 )}{∣ 2 x + 1 ∣}

Schritte zur Lösung

\frac{d}{d x} (∣ 2 x + 1 ∣)

Wende die Kettenregel an:

\frac{2 x + 1}{∣ ( 2 x + 1 ) ∣} \frac{d}{d x} (2 x + 1)

= \frac{2 x + 1}{∣ ( 2 x + 1 ) ∣} \frac{d}{d x} (2 x + 1)

\frac{d}{d x} (2 x + 1) = 2

= \frac{2 x + 1}{∣ ( 2 x + 1 ) ∣} \cdot 2

Multipliziere Brüche:

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

= \frac{2 ( 2 x + 1 )}{∣ 2 x + 1 ∣}

\int_{1}^{8} \frac{2}{x ^{3}} d x

\frac{\partial}{\partial x} ((7 x)^{4})

x \to 0 + lim ((- ln (x))^{x})

\frac{d y}{d x} = \frac{( 3 x ^{2} )}{y ^{2}}

\frac{d}{d x} (x - sin (2 x))