derivative f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1

解

導関数

f (x) = 4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1

2 x + \frac{16 x ^{\frac{1}{3}}}{3} - x^{\frac{1}{2}} - 4

解答ステップ

\frac{d}{d x} (4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1)

簡素化

4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1 : 4 x^{\frac{4}{3}} - \frac{2 x ^{\frac{3}{2}}}{3} + x^{2} - 4 x + 1

= \frac{d}{d x} (4 x^{\frac{4}{3}} - \frac{2 x ^{\frac{3}{2}}}{3} + x^{2} - 4 x + 1)

和/差の法則を適用:

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

= \frac{d}{d x} (4 x^{\frac{4}{3}}) - \frac{d}{d x} (\frac{2 x ^{\frac{3}{2}}}{3}) + \frac{d}{d x} (x^{2}) - \frac{d}{d x} (4 x) + \frac{d}{d x} (1)

\frac{d}{d x} (4 x^{\frac{4}{3}}) = \frac{16 x ^{\frac{1}{3}}}{3}

\frac{d}{d x} (\frac{2 x ^{\frac{3}{2}}}{3}) = x^{\frac{1}{2}}

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

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

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

= \frac{16 x ^{\frac{1}{3}}}{3} - x^{\frac{1}{2}} + 2 x - 4 + 0

簡素化

= 2 x + \frac{16 x ^{\frac{1}{3}}}{3} - x^{\frac{1}{2}} - 4