d/(dt)((x)(2x^2+1))

Lösung

\frac{d}{d t} ((t) (2 t^{2} + 1))

6 t^{2} + 1

Schritte zur Lösung

\frac{d}{d t} ((t) (2 t^{2} + 1))

Wende die Produktregel an:

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

f = t, g = 2 t^{2} + 1

= \frac{d t}{d t} (2 t^{2} + 1) + \frac{d}{d t} (2 t^{2} + 1) t

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

\frac{d}{d t} (2 t^{2} + 1) = 4 t

= 1 \cdot (2 t^{2} + 1) + 4 tt

Vereinfache

1 \cdot (2 t^{2} + 1) + 4 tt : 6 t^{2} + 1

= 6 t^{2} + 1

e x p an d \frac{2 ( 216 - 216 x + 72 x ^{2} - 8 x ^{3} )}{27}

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