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

Lösung

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

\frac{x ^{6} + 1}{x ^{2} ( x ^{6} - 1 ) ^{\frac{2}{3}}}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

\frac{1}{3 ( x ^{3} - \frac{1}{x ^{3}} ) ^{\frac{2}{3}}} (3 x^{2} + \frac{3}{x ^{4}}) : \frac{x ^{6} + 1}{x ^{2} ( x ^{6} - 1 ) ^{\frac{2}{3}}}

= \frac{x ^{6} + 1}{x ^{2} ( x ^{6} - 1 ) ^{\frac{2}{3}}}

l a pl a ce t r an s f or m e^{- 4 t} \cdot cos (3 t)

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