d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)})

Lösung

\frac{d}{d t} (4 \frac{e ^{5 t}}{t ^{2} + 1})

\frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

Schritte zur Lösung

\frac{d}{d t} (4 \frac{e ^{5 t}}{t ^{2} + 1})

Wende die Kettenregel an:

\frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1})

= \frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1})

\frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1}) = \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}}

= \frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \cdot \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}}

Vereinfache

\frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \cdot \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}} : \frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

= \frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

\frac{y ^{^{''}}}{y} = - 3

\int - 25 sin (- 5 t) d t

\frac{d}{d x} (arccos (e^{6 x}))

x \to - 2 - lim (x + 2)

\frac{d y}{d x} = x^{2} e^{- 4 x} - 4 y