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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

d er i v a t i v e arccot (3 x)

n = 1 \sum \infty \frac{n !}{1 4 ^{n}}

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

n = 0 \sum \infty \frac{( 8 ^{n} )}{n !}

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