derivative f(x)=-3/2 x^3+2/5 x^2-4

解

導関数

f (x) = - \frac{3}{2} x^{3} + \frac{2}{5} x^{2} - 4

- \frac{9 x ^{2}}{2} + \frac{4 x}{5}

解答ステップ

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

簡素化

- \frac{3}{2} x^{3} + \frac{2}{5} x^{2} - 4 : - \frac{3 x ^{3}}{2} + \frac{2 x ^{2}}{5} - 4

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

和/差の法則を適用:

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

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

\frac{d}{d x} (\frac{3 x ^{3}}{2}) = \frac{9 x ^{2}}{2}

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

\frac{d}{d x} (4) = 0

= - \frac{9 x ^{2}}{2} + \frac{4 x}{5} - 0

簡素化

= - \frac{9 x ^{2}}{2} + \frac{4 x}{5}