d/(dt)(\sqrt[5]{t}+4sqrt(t^5))

Lösung

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

\frac{1 + 50 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d t} (5 t) = \frac{1}{5 t ^{\frac{4}{5}}}

\frac{d}{d t} (4 t^{5}) = 10 t^{\frac{3}{2}}

= \frac{1}{5 t ^{\frac{4}{5}}} + 10 t^{\frac{3}{2}}

Vereinfache

\frac{1}{5 t ^{\frac{4}{5}}} + 10 t^{\frac{3}{2}} : \frac{1 + 50 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

= \frac{1 + 50 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

\frac{d}{d x} (\frac{2}{3 x})

y^{^{'''}} - 2 y^{^{'}} + y = 0

\int - 9 sin (3 x) d x

\int 12 x ln (3 x) d x

\frac{d y}{d t} = t y^{\frac{1}{3}}