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

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

- \frac{1}{( 1 + t ^{2} ) ^{2}} \cdot 2 t : - \frac{2 t}{( 1 + t ^{2} ) ^{2}}

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

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

in v erse l a pl a ce \frac{6}{s} + \frac{24}{s ^{2} + 16}

x \to 6 lim (\frac{3}{x})

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + 2})

\int x sec^{2} (2 x) d x