d/(dt)((s)(9+0.02s))

Lösung

\frac{d}{d t} ((t) (9 + 0.02 t))

9 + 0.04 t

Schritte zur Lösung

\frac{d}{d t} ((t) (9 + 0.02 t))

Wende die Produktregel an:

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

f = t, g = 9 + 0.02 t

= \frac{d t}{d t} (9 + 0.02 t) + \frac{d}{d t} (9 + 0.02 t) t

\frac{d t}{d t} = 1

\frac{d}{d t} (9 + 0.02 t) = 0.02

= 1 \cdot (9 + 0.02 t) + 0.02 t

Vereinfache

= 9 + 0.04 t

e x p an d (3 + 2) (3 + 2)

s im pl i f y 2 + \frac{1}{12} (8.1 - 8) - \frac{\frac{1}{144} ( 8.1 - 8 ) ^{2}}{2}

e x p an d \frac{3 s + 7}{s ( s + 2 )}

e x p an d \frac{e ^{- s}}{s ( s - 1 )}

e x p an d, 6 x^{2} + 17 x + 5 (3 x + a) (b x + c)