derivative t^{11}ln|t|

解

導関数

t^{11} ln ∣ t ∣

11 t^{10} ln ∣ t ∣ + t^{10}

解答ステップ

\frac{d}{d t} (t^{11} ln ∣ t ∣)

積の法則を適用:

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

f = t^{11}, g = ln ∣ t ∣

= \frac{d}{d t} (t^{11}) ln ∣ t ∣ + \frac{d}{d t} (ln ∣ t ∣) t^{11}

\frac{d}{d t} (t^{11}) = 11 t^{10}

\frac{d}{d t} (ln ∣ t ∣) = \frac{1}{t}

= 11 t^{10} ln ∣ t ∣ + \frac{1}{t} t^{11}

\frac{1}{t} t^{11} = t^{10}

= 11 t^{10} ln ∣ t ∣ + t^{10}