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

Lösung

\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)

Schritte zur Lösung

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

Wende die Produktregel an:

(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)

e x p an d \frac{5 ( x - 2 )}{x - 1}

s im pl i f y \frac{z ( k z + 5 )}{z ^{4} + k z ^{2} + 5.1 z + 0.5} + 1

e x p an d \frac{( 5 x + 9 ) ( x + 3 )}{( x + 2 ) ( 4 x + 3 )}

s im pl i f y (m^{2} n^{2} - \frac{3}{4}) (m^{2} n^{2} + \frac{1}{2})

e x p an d sin^{2} (θ) (1 + cot (θ))