derivative f(t)=(9t+2)^{2/3}

解

導関数

f (t) = (9 t + 2)^{\frac{2}{3}}

\frac{6}{( 9 t + 2 ) ^{\frac{1}{3}}}

解答ステップ

\frac{d}{d t} ((9 t + 2)^{\frac{2}{3}})

連鎖法則を適用:

\frac{2}{3 ( 9 t + 2 ) ^{\frac{1}{3}}} \frac{d}{d t} (9 t + 2)

= \frac{2}{3 ( 9 t + 2 ) ^{\frac{1}{3}}} \frac{d}{d t} (9 t + 2)

\frac{d}{d t} (9 t + 2) = 9

= \frac{2}{3 ( 9 t + 2 ) ^{\frac{1}{3}}} \cdot 9

簡素化

\frac{2}{3 ( 9 t + 2 ) ^{\frac{1}{3}}} \cdot 9 : \frac{6}{( 9 t + 2 ) ^{\frac{1}{3}}}

= \frac{6}{( 9 t + 2 ) ^{\frac{1}{3}}}