derivative of sqrt(3x^2+1)(3x^4+1^3)

解

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

\frac{3 x ( 3 x ^{4} + 1 ) ^{3}}{3 x ^{2} + 1} + 36 x^{3} (3 x^{4} + 1)^{2} 3 x^{2} + 1

解答ステップ

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

積の法則を適用:

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

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

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

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

\frac{d}{d x} ((3 x^{4} + 1)^{3}) = 36 x^{3} (3 x^{4} + 1)^{2}

= \frac{3 x}{3 x ^{2} + 1} (3 x^{4} + 1)^{3} + 36 x^{3} (3 x^{4} + 1)^{2} 3 x^{2} + 1

分数を乗じる:

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

= \frac{3 x ( 3 x ^{4} + 1 ) ^{3}}{3 x ^{2} + 1} + 36 x^{3} (3 x^{4} + 1)^{2} 3 x^{2} + 1