derivative (2x-1)ln(2x-1)

解

導関数

(2 x - 1) ln (2 x - 1)

2 ln (2 x - 1) + 2

解答ステップ

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

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 2 x - 1, g = ln (2 x - 1)

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

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

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

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

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

= 2 ln (2 x - 1) + 2