derivative ((\sqrt[3]{t}))/(t-3)

Lösung

ableitung von

\frac{( 3 t )}{t - 3}

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

Schritte zur Lösung

\frac{d}{d t} (\frac{3 t}{t - 3})

Wende die Quotientenregel an:

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

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

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

\frac{d}{d t} (t - 3) = 1

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

Vereinfache

\frac{\frac{1}{3 t ^{\frac{2}{3}}} ( t - 3 ) - 1 \cdot 3 t}{( t - 3 ) ^{2}} : \frac{t - 3 - 3 3 t t ^{\frac{2}{3}}}{3 t ^{\frac{2}{3}} ( t - 3 ) ^{2}}

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

\int - \frac{1}{2} d x

\int_{0}^{1} x - x^{2} d x

\int (y - 1) (2 y + 1) d y

\frac{d y}{d x} + 2 x y = 2 x^{3}

in v erse l a pl a ce \frac{3 ( s + 3 )}{s ^{2} + 6 s + 8}