derivative of log_{2x+1}(2x+3)

解

\frac{d}{d x} (lo g_{2 x + 1} (2 x + 3))

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

解答ステップ

\frac{d}{d x} (lo g_{2 x + 1} (2 x + 3))

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 ( 2 x + 3 )}{ln ( 2 x + 1 )})

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

簡素化

\frac{\frac{2}{2 x + 3} ln ( 2 x + 1 ) - \frac{2}{2 x + 1} ln ( 2 x + 3 )}{ln ^{2} ( 2 x + 1 )} : \frac{2 ln ( 2 x + 1 ) ( 2 x + 1 ) - 2 ln ( 2 x + 3 ) ( 2 x + 3 )}{ln ^{2} ( 2 x + 1 ) ( 2 x + 3 ) ( 2 x + 1 )}

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