derivative y=(2x-1)^2ln(2x-1)

解

導関数

y = (2 x - 1)^{2} ln (2 x - 1)

4 ln (2 x - 1) (2 x - 1) + 2 (2 x - 1)

解答ステップ

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

積の法則を適用:

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

f = (2 x - 1)^{2}, g = ln (2 x - 1)

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

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

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

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

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

= 4 ln (2 x - 1) (2 x - 1) + 2 (2 x - 1)