derivative y=ln(3x+1)e^{x^2+4}

解

導関数

y = ln (3 x + 1) e^{x^{2} + 4}

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

解答ステップ

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

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = ln (3 x + 1), g = e^{x^{2} + 4}

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

\frac{d}{d x} (ln (3 x + 1)) = \frac{3}{3 x + 1}

\frac{d}{d x} (e^{x^{2} + 4}) = e^{x^{2} + 4} \cdot 2 x

= \frac{3}{3 x + 1} e^{x^{2} + 4} + e^{x^{2} + 4} \cdot 2 x ln (3 x + 1)

分数を乗じる:

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

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