derivative f(x)=ax^3+bx^2+cx+d

解

導関数

f (x) = a x^{3} + b x^{2} + c x + d

3 a x^{2} + 2 b x + c

解答ステップ

\frac{d}{d x} (a x^{3} + b x^{2} + c x + d)

和/差の法則を適用:

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

= \frac{d}{d x} (a x^{3}) + \frac{d}{d x} (b x^{2}) + \frac{d}{d x} (c x) + \frac{dd}{d x}

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

\frac{d}{d x} (b x^{2}) = 2 b x

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

\frac{dd}{d x} = 0

= 3 a x^{2} + 2 b x + c + 0

簡素化

= 3 a x^{2} + 2 b x + c