(\partial)/(\partial x)(ln((2x-1)/(x-1)))

解

\frac{\partial}{\partial x} (ln (\frac{2 x - 1}{x - 1}))

- \frac{1}{( 2 x - 1 ) ( x - 1 )}

解答ステップ

\frac{\partial}{\partial x} (ln (\frac{2 x - 1}{x - 1}))

連鎖法則を適用:

\frac{1}{\frac{2 x - 1}{x - 1}} \frac{\partial}{\partial x} (\frac{2 x - 1}{x - 1})

= \frac{1}{\frac{2 x - 1}{x - 1}} \frac{\partial}{\partial x} (\frac{2 x - 1}{x - 1})

\frac{\partial}{\partial x} (\frac{2 x - 1}{x - 1}) = - \frac{1}{( x - 1 ) ^{2}}

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

簡素化

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

= - \frac{1}{( 2 x - 1 ) ( x - 1 )}