derivative of ln(3x^3+4x^2+x+4)

解

\frac{d}{d x} (ln (3 x^{3} + 4 x^{2} + x + 4))

\frac{9 x ^{2} + 8 x + 1}{3 x ^{3} + 4 x ^{2} + x + 4}

解答ステップ

\frac{d}{d x} (ln (3 x^{3} + 4 x^{2} + x + 4))

連鎖法則を適用:

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

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

\frac{d}{d x} (3 x^{3} + 4 x^{2} + x + 4) = 9 x^{2} + 8 x + 1

= \frac{1}{3 x ^{3} + 4 x ^{2} + x + 4} (9 x^{2} + 8 x + 1)

簡素化

\frac{1}{3 x ^{3} + 4 x ^{2} + x + 4} (9 x^{2} + 8 x + 1) : \frac{9 x ^{2} + 8 x + 1}{3 x ^{3} + 4 x ^{2} + x + 4}

= \frac{9 x ^{2} + 8 x + 1}{3 x ^{3} + 4 x ^{2} + x + 4}