derivative of log_{2}(2x+1)

Lösung

\frac{d}{d x} (lo g_{2} (2 x + 1))

\frac{2}{ln ( 2 ) ( 2 x + 1 )}

Schritte zur Lösung

\frac{d}{d x} (lo g_{2} (2 x + 1))

Wende die log Regel an:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

\frac{1}{ln ( 2 )} \cdot \frac{1}{2 x + 1} \cdot 2 : \frac{2}{ln ( 2 ) ( 2 x + 1 )}

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

a re a x^{2} + 2, x + 2

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