derivative ln(6x^3+3x^2+3x+3)

解

導関数

ln (6 x^{3} + 3 x^{2} + 3 x + 3)

\frac{6 x ^{2} + 2 x + 1}{2 x ^{3} + x ^{2} + x + 1}

解答ステップ

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

連鎖法則を適用:

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

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

\frac{d}{d x} (6 x^{3} + 3 x^{2} + 3 x + 3) = 18 x^{2} + 6 x + 3

= \frac{1}{6 x ^{3} + 3 x ^{2} + 3 x + 3} (18 x^{2} + 6 x + 3)

簡素化

\frac{1}{6 x ^{3} + 3 x ^{2} + 3 x + 3} (18 x^{2} + 6 x + 3) : \frac{6 x ^{2} + 2 x + 1}{2 x ^{3} + x ^{2} + x + 1}

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