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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

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

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

a re a x^{2}, 1, 0

t an g e n t f (x) = 9 x^{2} + 7 x - 6

\int \frac{1}{t ^{2} 169 - t ^{2}} d t

n or ma l y = \frac{x}{x + 6}, (4, 0.2)

\int_{1}^{e} ln (12 x) d x