derivative f(x)= 1/5 x^4-4x^2+1

解

導関数

f (x) = \frac{1}{5} x^{4} - 4 x^{2} + 1

\frac{4 x ^{3}}{5} - 8 x

解答ステップ

\frac{d}{d x} (\frac{1}{5} x^{4} - 4 x^{2} + 1)

簡素化

\frac{1}{5} x^{4} - 4 x^{2} + 1 : \frac{x ^{4}}{5} - 4 x^{2} + 1

= \frac{d}{d x} (\frac{x ^{4}}{5} - 4 x^{2} + 1)

和/差の法則を適用:

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

= \frac{d}{d x} (\frac{x ^{4}}{5}) - \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (1)

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

\frac{d}{d x} (4 x^{2}) = 8 x

\frac{d}{d x} (1) = 0

= \frac{4 x ^{3}}{5} - 8 x + 0

簡素化

= \frac{4 x ^{3}}{5} - 8 x