derivative f(x)=x^5-5x^3+x-1

解

導関数

f (x) = x^{5} - 5 x^{3} + x - 1

5 x^{4} - 15 x^{2} + 1

解答ステップ

\frac{d}{d x} (x^{5} - 5 x^{3} + x - 1)

和/差の法則を適用:

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

= \frac{d}{d x} (x^{5}) - \frac{d}{d x} (5 x^{3}) + \frac{d x}{d x} - \frac{d}{d x} (1)

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

\frac{d}{d x} (5 x^{3}) = 15 x^{2}

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

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

= 5 x^{4} - 15 x^{2} + 1 - 0

簡素化

= 5 x^{4} - 15 x^{2} + 1