derivative (\sqrt[3]{t})/(t-9)

Lösung

ableitung von

\frac{3 t}{t - 9}

\frac{t - 9 - 3 3 t t ^{\frac{2}{3}}}{3 t ^{\frac{2}{3}} ( t - 9 ) ^{2}}

Schritte zur Lösung

\frac{d}{d t} (\frac{3 t}{t - 9})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d t} ( 3 t ) ( t - 9 ) - \frac{d}{d t} ( t - 9 ) 3 t}{( t - 9 ) ^{2}}

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

\frac{d}{d t} (t - 9) = 1

= \frac{\frac{1}{3 t ^{\frac{2}{3}}} ( t - 9 ) - 1 \cdot 3 t}{( t - 9 ) ^{2}}

Vereinfache

\frac{\frac{1}{3 t ^{\frac{2}{3}}} ( t - 9 ) - 1 \cdot 3 t}{( t - 9 ) ^{2}} : \frac{t - 9 - 3 3 t t ^{\frac{2}{3}}}{3 t ^{\frac{2}{3}} ( t - 9 ) ^{2}}

= \frac{t - 9 - 3 3 t t ^{\frac{2}{3}}}{3 t ^{\frac{2}{3}} ( t - 9 ) ^{2}}

x \to 1 lim (\frac{x + 2}{x ^{3} + 8})

\frac{\partial}{\partial x} (x^{4} ln (yz))

n = 1 \sum \infty \frac{1}{1 0 ^{n}}

n = 1 \sum \infty (\frac{2}{7}) \cdot n

\int 2 t + 2 d t