derivative of 4x^5+2x^3+6x^2+3x

解

\frac{d}{d x} (4 x^{5} + 2 x^{3} + 6 x^{2} + 3 x)

20 x^{4} + 6 x^{2} + 12 x + 3

解答ステップ

\frac{d}{d x} (4 x^{5} + 2 x^{3} + 6 x^{2} + 3 x)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (4 x^{5}) + \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (3 x)

\frac{d}{d x} (4 x^{5}) = 20 x^{4}

\frac{d}{d x} (2 x^{3}) = 6 x^{2}

\frac{d}{d x} (6 x^{2}) = 12 x

\frac{d}{d x} (3 x) = 3

= 20 x^{4} + 6 x^{2} + 12 x + 3