derivative f(t)=(2t-1)^7+14t(2t-1)^6

解

導関数

f (t) = (2 t - 1)^{7} + 14 t (2 t - 1)^{6}

7168 t^{6} - 18816 t^{5} + 20160 t^{4} - 11200 t^{3} + 3360 t^{2} - 504 t + 28

解答ステップ

\frac{d}{d t} ((2 t - 1)^{7} + 14 t (2 t - 1)^{6})

簡素化

(2 t - 1)^{7} + 14 t (2 t - 1)^{6} : 1024 t^{7} - 3136 t^{6} + 4032 t^{5} - 2800 t^{4} + 1120 t^{3} - 252 t^{2} + 28 t - 1

= \frac{d}{d t} (1024 t^{7} - 3136 t^{6} + 4032 t^{5} - 2800 t^{4} + 1120 t^{3} - 252 t^{2} + 28 t - 1)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (1024 t^{7}) - \frac{d}{d t} (3136 t^{6}) + \frac{d}{d t} (4032 t^{5}) - \frac{d}{d t} (2800 t^{4}) + \frac{d}{d t} (1120 t^{3}) - \frac{d}{d t} (252 t^{2}) + \frac{d}{d t} (28 t) - \frac{d}{d t} (1)

\frac{d}{d t} (1024 t^{7}) = 7168 t^{6}

\frac{d}{d t} (3136 t^{6}) = 18816 t^{5}

\frac{d}{d t} (4032 t^{5}) = 20160 t^{4}

\frac{d}{d t} (2800 t^{4}) = 11200 t^{3}

\frac{d}{d t} (1120 t^{3}) = 3360 t^{2}

\frac{d}{d t} (252 t^{2}) = 504 t

\frac{d}{d t} (28 t) = 28

\frac{d}{d t} (1) = 0

= 7168 t^{6} - 18816 t^{5} + 20160 t^{4} - 11200 t^{3} + 3360 t^{2} - 504 t + 28 - 0

簡素化

= 7168 t^{6} - 18816 t^{5} + 20160 t^{4} - 11200 t^{3} + 3360 t^{2} - 504 t + 28