derivative of log_{6}(5x+1)

解

\frac{d}{d x} (lo g_{6} (5 x + 1))

\frac{5}{ln ( 6 ) ( 5 x + 1 )}

解答ステップ

\frac{d}{d x} (lo g_{6} (5 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 ( 5 x + 1 )}{ln ( 6 )})

定数を除去:

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

= \frac{1}{ln ( 6 )} \frac{d}{d x} (ln (5 x + 1))

連鎖法則を適用:

\frac{1}{5 x + 1} \frac{d}{d x} (5 x + 1)

= \frac{1}{5 x + 1} \frac{d}{d x} (5 x + 1)

\frac{d}{d x} (5 x + 1) = 5

= \frac{1}{ln ( 6 )} \cdot \frac{1}{5 x + 1} \cdot 5

簡素化

\frac{1}{ln ( 6 )} \cdot \frac{1}{5 x + 1} \cdot 5 : \frac{5}{ln ( 6 ) ( 5 x + 1 )}

= \frac{5}{ln ( 6 ) ( 5 x + 1 )}