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

Lösung

ableitung von

\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}}

t a y l or (1 - x)^{\frac{1}{4}}

n = 1 \sum \infty (\frac{1}{8})^{n - 1}

\frac{d}{d x} (\frac{c e ^{x}}{1 + c e ^{x}})

t an g e n t y = x^{2} + 7, (3, 16)

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