derivative ((t^3+t))/(t^4-2)

Lösung

ableitung von

\frac{( t ^{3} + t )}{t ^{4} - 2}

\frac{- t ^{6} - 3 t ^{4} - 6 t ^{2} - 2}{( t ^{4} - 2 ) ^{2}}

Schritte zur Lösung

\frac{d}{d t} (\frac{t ^{3} + t}{t ^{4} - 2})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d t} ( t ^{3} + t ) ( t ^{4} - 2 ) - \frac{d}{d t} ( t ^{4} - 2 ) ( t ^{3} + t )}{( t ^{4} - 2 ) ^{2}}

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

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

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

Multipliziere aus

(3 t^{2} + 1) (t^{4} - 2) - 4 t^{3} (t^{3} + t) : - t^{6} - 3 t^{4} - 6 t^{2} - 2

= \frac{- t ^{6} - 3 t ^{4} - 6 t ^{2} - 2}{( t ^{4} - 2 ) ^{2}}

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

\int \frac{21}{4 + 25 x ^{2}} d x

n \to \infty lim (e^{n})

in v erse l a pl a ce \frac{s + 5}{( s + 1 ) ( s + 3 )}

\frac{d}{d x} (\frac{12}{x})