derivative of ln|1+x|

Lösung

\frac{d}{d x} (ln ∣ 1 + x ∣)

\frac{1}{1 + x}

Schritte zur Lösung

\frac{d}{d x} (ln ∣ 1 + x ∣)

Wende die Kettenregel an:

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

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

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

= \frac{1}{∣ 1 + x ∣} \cdot \frac{1 + x}{∣ 1 + x ∣}

Vereinfache

\frac{1}{∣ 1 + x ∣} \cdot \frac{1 + x}{∣ 1 + x ∣} : \frac{1}{1 + x}

= \frac{1}{1 + x}

d er i v a t i v e arcsec (x) - 8 x

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