derivative of \sqrt[3]{x^3+6x+7}

解

\frac{d}{d x} (3 x^{3} + 6 x + 7)

\frac{x ^{2} + 2}{( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}}

解答ステップ

\frac{d}{d x} (3 x^{3} + 6 x + 7)

連鎖法則を適用:

\frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{3} + 6 x + 7)

= \frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{3} + 6 x + 7)

\frac{d}{d x} (x^{3} + 6 x + 7) = 3 x^{2} + 6

= \frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} (3 x^{2} + 6)

簡素化

\frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} (3 x^{2} + 6) : \frac{x ^{2} + 2}{( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}}

= \frac{x ^{2} + 2}{( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}}