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

解

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

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

解答ステップ

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

連鎖法則を適用:

\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)

簡素化

\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}}}