derivative of |x^3+3x^2-8|

解

\frac{d}{d x} (x^{3} + 3 x^{2} - 8)

\frac{( x ^{3} + 3 x ^{2} - 8 ) ( 3 x ^{2} + 6 x )}{∣ x ^{3} + 3 x ^{2} - 8 ∣}

解答ステップ

\frac{d}{d x} (x^{3} + 3 x^{2} - 8)

連鎖法則を適用:

\frac{x ^{3} + 3 x ^{2} - 8}{∣ ( x ^{3} + 3 x ^{2} - 8 ) ∣} \frac{d}{d x} (x^{3} + 3 x^{2} - 8)

= \frac{x ^{3} + 3 x ^{2} - 8}{∣ ( x ^{3} + 3 x ^{2} - 8 ) ∣} \frac{d}{d x} (x^{3} + 3 x^{2} - 8)

\frac{d}{d x} (x^{3} + 3 x^{2} - 8) = 3 x^{2} + 6 x

= \frac{x ^{3} + 3 x ^{2} - 8}{∣ ( x ^{3} + 3 x ^{2} - 8 ) ∣} (3 x^{2} + 6 x)

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( x ^{3} + 3 x ^{2} - 8 ) ( 3 x ^{2} + 6 x )}{∣ x ^{3} + 3 x ^{2} - 8 ∣}