derivative 49+23sin((2pi)/(52)(t-10))

解

導関数

49 + 23 sin (\frac{2 π}{52} (t - 10))

\frac{23 π}{26} cos (\frac{π ( t - 10 )}{26})

解答ステップ

\frac{d}{d t} (49 + 23 sin (\frac{2 π}{52} (t - 10)))

簡素化

49 + 23 sin (\frac{2 π}{52} (t - 10)) : 49 + 23 sin (\frac{π ( t - 10 )}{26})

= \frac{d}{d t} (49 + 23 sin (\frac{π ( t - 10 )}{26}))

和/差の法則を適用:

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

= \frac{d}{d t} (49) + \frac{d}{d t} (23 sin (\frac{π ( t - 10 )}{26}))

\frac{d}{d t} (49) = 0

\frac{d}{d t} (23 sin (\frac{π ( t - 10 )}{26})) = \frac{23 π}{26} cos (\frac{π ( t - 10 )}{26})

= 0 + \frac{23 π}{26} cos (\frac{π ( t - 10 )}{26})

簡素化

= \frac{23 π}{26} cos (\frac{π ( t - 10 )}{26})