derivative of log_{3}(8x-1)

解

\frac{d}{d x} (lo g_{3} (8 x - 1))

\frac{8}{ln ( 3 ) ( 8 x - 1 )}

解答ステップ

\frac{d}{d x} (lo g_{3} (8 x - 1))

対数の規則を適用する:

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

= \frac{d}{d x} (\frac{ln ( 8 x - 1 )}{ln ( 3 )})

定数を除去:

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

= \frac{1}{ln ( 3 )} \frac{d}{d x} (ln (8 x - 1))

連鎖法則を適用:

\frac{1}{8 x - 1} \frac{d}{d x} (8 x - 1)

= \frac{1}{8 x - 1} \frac{d}{d x} (8 x - 1)

\frac{d}{d x} (8 x - 1) = 8

= \frac{1}{ln ( 3 )} \cdot \frac{1}{8 x - 1} \cdot 8

簡素化

\frac{1}{ln ( 3 )} \cdot \frac{1}{8 x - 1} \cdot 8 : \frac{8}{ln ( 3 ) ( 8 x - 1 )}

= \frac{8}{ln ( 3 ) ( 8 x - 1 )}