derivative f(t)=(t^{62}+3t+3)(5t^2+4)

Lösung

ableitung von

f (t) = (t^{62} + 3 t + 3) (5 t^{2} + 4)

320 t^{63} + 248 t^{61} + 45 t^{2} + 30 t + 12

Schritte zur Lösung

\frac{d}{d t} ((t^{62} + 3 t + 3) (5 t^{2} + 4))

Wende die Produktregel an:

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

f = t^{62} + 3 t + 3, g = 5 t^{2} + 4

= \frac{d}{d t} (t^{62} + 3 t + 3) (5 t^{2} + 4) + \frac{d}{d t} (5 t^{2} + 4) (t^{62} + 3 t + 3)

\frac{d}{d t} (t^{62} + 3 t + 3) = 62 t^{61} + 3

\frac{d}{d t} (5 t^{2} + 4) = 10 t

= (62 t^{61} + 3) (5 t^{2} + 4) + 10 t (t^{62} + 3 t + 3)

Vereinfache

(62 t^{61} + 3) (5 t^{2} + 4) + 10 t (t^{62} + 3 t + 3) : 320 t^{63} + 248 t^{61} + 45 t^{2} + 30 t + 12

= 320 t^{63} + 248 t^{61} + 45 t^{2} + 30 t + 12

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