derivative of ln(sqrt(x^2e^{x^{2+1)}})

解

\frac{d}{d x} (ln (x^{2} e^{x^{2 + 1}}))

\frac{3 x ^{2} ∣ x ∣ + 2}{2 ∣ x ∣}

解答ステップ

\frac{d}{d x} (ln (x^{2} e^{x^{2 + 1}}))

簡素化

ln (x^{2} e^{x^{2 + 1}}) : ln (x^{2} e^{x^{3}})

= \frac{d}{d x} (ln (x^{2} e^{x^{3}}))

連鎖法則を適用:

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

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

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

= \frac{1}{x ^{2} e ^{x^{3}}} (\frac{x e ^{x^{3}}}{∣ x ∣} + \frac{3 e ^{x^{3}} x ^{2} x ^{2}}{2 e ^{x^{3}}})

簡素化

\frac{1}{x ^{2} e ^{x^{3}}} (\frac{x e ^{x^{3}}}{∣ x ∣} + \frac{3 e ^{x^{3}} x ^{2} x ^{2}}{2 e ^{x^{3}}}) : \frac{3 x ^{2} ∣ x ∣ + 2}{2 ∣ x ∣}

= \frac{3 x ^{2} ∣ x ∣ + 2}{2 ∣ x ∣}