derivative of log_{3}(\sqrt[5]{x+2})

解

\frac{d}{d x} (lo g_{3} (5 x + 2))

\frac{1}{5 ln ( 3 ) 5 x + 2 ( x + 2 ) ^{\frac{4}{5}}}

解答ステップ

\frac{d}{d x} (lo g_{3} (5 x + 2))

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 + 2 )}{ln ( 3 )})

定数を除去:

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

= \frac{1}{ln ( 3 )} \frac{d}{d x} (ln (5 x + 2))

連鎖法則を適用:

\frac{1}{5 x + 2} \frac{d}{d x} (5 x + 2)

= \frac{1}{5 x + 2} \frac{d}{d x} (5 x + 2)

\frac{d}{d x} (5 x + 2) = \frac{1}{5 ( x + 2 ) ^{\frac{4}{5}}}

= \frac{1}{ln ( 3 )} \cdot \frac{1}{5 x + 2} \cdot \frac{1}{5 ( x + 2 ) ^{\frac{4}{5}}}

簡素化

\frac{1}{ln ( 3 )} \cdot \frac{1}{5 x + 2} \cdot \frac{1}{5 ( x + 2 ) ^{\frac{4}{5}}} : \frac{1}{5 ln ( 3 ) 5 x + 2 ( x + 2 ) ^{\frac{4}{5}}}

= \frac{1}{5 ln ( 3 ) 5 x + 2 ( x + 2 ) ^{\frac{4}{5}}}