derivative 5(x^{2n-1})*(2x^{3n+1})^2

解

導関数

5 (x^{2 n - 1}) \cdot (2 x^{3 n + 1})^{2}

5 (x^{2 n - 2} (2 n - 1) (2 x^{3 n + 1})^{2} + 8 x^{6 n + 1} (3 n + 1) x^{2 n - 1})

解答ステップ

\frac{d}{d x} (5 x^{2 n - 1} (2 x^{3 n + 1})^{2})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d x} (x^{2 n - 1} (2 x^{3 n + 1})^{2})

積の法則を適用:

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

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

= 5 (\frac{d}{d x} (x^{2 n - 1}) (2 x^{3 n + 1})^{2} + \frac{d}{d x} ((2 x^{3 n + 1})^{2}) x^{2 n - 1})

\frac{d}{d x} (x^{2 n - 1}) = x^{2 n - 2} (2 n - 1)

\frac{d}{d x} ((2 x^{3 n + 1})^{2}) = 8 x^{6 n + 1} (3 n + 1)

= 5 (x^{2 n - 2} (2 n - 1) (2 x^{3 n + 1})^{2} + 8 x^{6 n + 1} (3 n + 1) x^{2 n - 1})