derivative f(x)=-8xsin(4x^2+1)

解

導関数

f (x) = - 8 x sin (4 x^{2} + 1)

- 8 (sin (4 x^{2} + 1) + 8 x^{2} cos (4 x^{2} + 1))

解答ステップ

\frac{d}{d x} (- 8 x sin (4 x^{2} + 1))

定数を除去:

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

= - 8 \frac{d}{d x} (x sin (4 x^{2} + 1))

積の法則を適用:

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

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

= - 8 (\frac{d x}{d x} sin (4 x^{2} + 1) + \frac{d}{d x} (sin (4 x^{2} + 1)) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (sin (4 x^{2} + 1)) = cos (4 x^{2} + 1) \cdot 8 x

= - 8 (1 \cdot sin (4 x^{2} + 1) + cos (4 x^{2} + 1) \cdot 8 xx)

簡素化

- 8 (1 \cdot sin (4 x^{2} + 1) + cos (4 x^{2} + 1) \cdot 8 xx) : - 8 (sin (4 x^{2} + 1) + 8 x^{2} cos (4 x^{2} + 1))

= - 8 (sin (4 x^{2} + 1) + 8 x^{2} cos (4 x^{2} + 1))