展開する (f(x)+g(x))(f(x)(x)x)'

解

展開する

(f (x) + g (x)) (f (x) (x) (x) x)^{'}

(x^{2} f (x) (x))' f (x) + (x^{2} f (x) (x))' g (x)

解答ステップ

(f (x) + g (x)) (f (x) (x) (x) x)^{'}

= (f (x) + g (x)) ((x) x f (x) (x))^{^{'}}

分配法則を適用する:

a (b + c) = ab + a c

a = (f (x) (x) (x) x)^{'}, b = f (x), c = g (x)

= (f (x) (x) (x) x)^{^{'}} f (x) + (f (x) (x) (x) x)^{^{'}} g (x)

= ((x) x f (x) (x))^{^{'}} f (x) + ((x) x f (x) (x))^{^{'}} g (x)

簡素化

((x) x f (x) (x))^{'} f (x) + ((x) x f (x) (x))^{'} g (x) : (x^{2} f (x) (x))^{'} f (x) + (x^{2} f (x) (x))^{'} g (x)

= (x^{2} f (x) (x))^{'} f (x) + (x^{2} f (x) (x))^{'} g (x)