derivative of (\sqrt[3]{x+1}/(3-4x))

Lösung

\frac{d}{d x} (\frac{3 x + 1}{3 - 4 x})

\frac{3 - 4 x + 12 3 x + 1 ( x + 1 ) ^{\frac{2}{3}}}{3 ( 3 - 4 x ) ^{2} ( x + 1 ) ^{\frac{2}{3}}}

Schritte zur Lösung

\frac{d}{d x} (\frac{3 x + 1}{3 - 4 x})

Wende die Quotientenregel an:

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

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

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

\frac{d}{d x} (3 - 4 x) = - 4

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

Vereinfache

\frac{\frac{1}{3 ( x + 1 ) ^{\frac{2}{3}}} ( 3 - 4 x ) - ( - 4 ) 3 x + 1}{( 3 - 4 x ) ^{2}} : \frac{3 - 4 x + 12 3 x + 1 ( x + 1 ) ^{\frac{2}{3}}}{3 ( 3 - 4 x ) ^{2} ( x + 1 ) ^{\frac{2}{3}}}

= \frac{3 - 4 x + 12 3 x + 1 ( x + 1 ) ^{\frac{2}{3}}}{3 ( 3 - 4 x ) ^{2} ( x + 1 ) ^{\frac{2}{3}}}

\int 64 x^{3} (4 x^{2} - 1) d x

x \to - 7 lim (h (x))

a re a y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{20}

\int_{- 3}^{- 1} \frac{2}{x} d x

x \to 5 lim (\frac{x + 3}{5 - x})