derivative g(x)=(5-x)^2(2x-1)^5

解

導関数

g (x) = (5 - x)^{2} (2 x - 1)^{5}

- 2 (5 - x) (2 x - 1)^{5} + 10 (2 x - 1)^{4} (5 - x)^{2}

解答ステップ

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

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (5 - x)^{2}, g = (2 x - 1)^{5}

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

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

\frac{d}{d x} ((2 x - 1)^{5}) = 10 (2 x - 1)^{4}

= (- 2 (5 - x)) (2 x - 1)^{5} + 10 (2 x - 1)^{4} (5 - x)^{2}

簡素化

= - 2 (5 - x) (2 x - 1)^{5} + 10 (2 x - 1)^{4} (5 - x)^{2}