d/(dt)(({f}(t))(2e^{(t^3)}+2cos(3t)))

解

\frac{d}{d t} ((f (t)) (2 e^{(t^{3})} + 2 cos (3 t)))

\frac{d}{d t} (f (t)) (2 e^{t^{3}} + 2 cos (3 t)) + (6 e^{t^{3}} t^{2} - 6 sin (3 t)) f (t)

解答ステップ

\frac{d}{d t} ((f (t)) (2 e^{(t^{3})} + 2 cos (3 t)))

積の法則を適用:

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

f = f (t), g = 2 e^{t^{3}} + 2 cos (3 t)

= \frac{d}{d t} (f (t)) (2 e^{t^{3}} + 2 cos (3 t)) + \frac{d}{d t} (2 e^{t^{3}} + 2 cos (3 t)) f (t)

\frac{d}{d t} (2 e^{t^{3}} + 2 cos (3 t)) = 6 e^{t^{3}} t^{2} - 6 sin (3 t)

= \frac{d}{d t} (f (t)) (2 e^{t^{3}} + 2 cos (3 t)) + (6 e^{t^{3}} t^{2} - 6 sin (3 t)) f (t)