derivative f(x)=(x-1)*ln(x-1)

解

導関数

f (x) = (x - 1) \cdot ln (x - 1)

ln (x - 1) + 1

解答ステップ

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

積の法則を適用:

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

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

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

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

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

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

簡素化

1 \cdot ln (x - 1) + \frac{1}{x - 1} (x - 1) : ln (x - 1) + 1

= ln (x - 1) + 1