d/(dt)(\sqrt[3]{t}(t^2+4))

Lösung

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

\frac{t ^{2} + 4 + 6 3 t t ^{\frac{5}{3}}}{3 t ^{\frac{2}{3}}}

Schritte zur Lösung

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

Wende die Produktregel an:

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

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

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

\frac{d}{d t} (3 t) = \frac{1}{3 t ^{\frac{2}{3}}}

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

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

Vereinfache

\frac{1}{3 t ^{\frac{2}{3}}} (t^{2} + 4) + 2 t 3 t : \frac{t ^{2} + 4 + 6 3 t t ^{\frac{5}{3}}}{3 t ^{\frac{2}{3}}}

= \frac{t ^{2} + 4 + 6 3 t t ^{\frac{5}{3}}}{3 t ^{\frac{2}{3}}}

\int - \frac{4}{x ^{2}} d x

\frac{d y}{d x} = \frac{y}{y ^{2}}, y (0) = 2

n = 1 \sum \infty \frac{1}{2 + sin ( n )}

\frac{d}{d x} (e^{14 x})

\int \frac{x ^{2}}{3 1 + 2 x} d x