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

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\frac{d}{d t} (e^{4 t})

0.125 y^{^{''}} + 4 y^{^{'}} + 28 y = 0

\frac{d}{d x} (6 x^{3} cos (x))

4 x y^{^{'}} + 2 y = 10 x^{6}

\int \frac{x ^{3} + x ^{2}}{x ^{5}} d x