d/(dt)((x)(4(x^2+1)))

Lösung

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

4 (3 t^{2} + 1)

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Produktregel an:

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

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

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

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

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

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

Vereinfache

4 (1 \cdot (t^{2} + 1) + 2 tt) : 4 (3 t^{2} + 1)

= 4 (3 t^{2} + 1)

e x p an d - \frac{1}{- 2 e ^{- x} ( x + 1 ) + C}

e x p an d x^{2} + 1^{2}

e x p an d (x - 1) e^{3}

e x p an d (2 d t + 2 d w) \cdot (3 d t - d w)

e x p an d \frac{( n + 1 ) ( 2 n + 1 )}{18 n}