derivative of ln(((2x+1^3)/((3x-1)^4)))

解

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

\frac{6 ( - x - 3 )}{( 2 x + 1 ) ( 3 x - 1 )}

解答ステップ

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

連鎖法則を適用:

\frac{1}{\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}} \frac{d}{d x} (\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}})

= \frac{1}{\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}} \frac{d}{d x} (\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}})

\frac{d}{d x} (\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}) = \frac{6 ( 2 x + 1 ) ^{2} ( - x - 3 )}{( 3 x - 1 ) ^{5}}

= \frac{1}{\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}} \cdot \frac{6 ( 2 x + 1 ) ^{2} ( - x - 3 )}{( 3 x - 1 ) ^{5}}

簡素化

\frac{1}{\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}} \cdot \frac{6 ( 2 x + 1 ) ^{2} ( - x - 3 )}{( 3 x - 1 ) ^{5}} : \frac{6 ( - x - 3 )}{( 2 x + 1 ) ( 3 x - 1 )}

= \frac{6 ( - x - 3 )}{( 2 x + 1 ) ( 3 x - 1 )}