derivative of \sqrt[3]{x^3-3x+c}

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\frac{d y}{d x} - 7 = 3 y - 7 x + 9

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