derivative y=t(ln(2t))^2

解

導関数

y = t (ln (2 t))^{2}

ln^{2} (2 t) + 2 ln (2 t)

解答ステップ

\frac{d}{d t} (t (ln (2 t))^{2})

積の法則を適用:

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

f = t, g = (ln (2 t))^{2}

= \frac{d t}{d t} (ln (2 t))^{2} + \frac{d}{d t} ((ln (2 t))^{2}) t

\frac{d t}{d t} = 1

\frac{d}{d t} ((ln (2 t))^{2}) = \frac{2 ln ( 2 t )}{t}

= 1 \cdot (ln (2 t))^{2} + \frac{2 ln ( 2 t )}{t} t

簡素化

1 \cdot ln^{2} (2 t) + \frac{2 ln ( 2 t )}{t} t : ln^{2} (2 t) + 2 ln (2 t)

= ln^{2} (2 t) + 2 ln (2 t)