derivative f(t)=(\sqrt[3]{t})/(t-6)

Lösung

ableitung von

f (t) = \frac{3 t}{t - 6}

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

Schritte zur Lösung

\frac{d}{d t} (\frac{3 t}{t - 6})

Wende die Quotientenregel an:

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

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

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

\frac{d}{d t} (t - 6) = 1

= \frac{\frac{1}{3 t ^{\frac{2}{3}}} ( t - 6 ) - 1 \cdot 3 t}{( t - 6 ) ^{2}}

Vereinfache

\frac{\frac{1}{3 t ^{\frac{2}{3}}} ( t - 6 ) - 1 \cdot 3 t}{( t - 6 ) ^{2}} : \frac{t - 6 - 3 3 t t ^{\frac{2}{3}}}{3 t ^{\frac{2}{3}} ( t - 6 ) ^{2}}

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

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

\int_{2}^{5} e^{- x} d x

\frac{\partial}{\partial y} (3 y^{2} cos (4 x))

(4 + x) \frac{d y}{d x} + y = 4 + x

d er i v a t i v e f (x) = 3 x - ln (x^{4})