展開する f''(x)(2x+1)^5

解

展開する

f^{''} (x) (2 x + 1)^{5}

32 x^{5} f'' (x) + 80 x^{4} f'' (x) + 80 x^{3} f'' (x) + 40 x^{2} f'' (x) + 10 x f'' (x) + f'' (x)

解答ステップ

f^{''} (x) (2 x + 1)^{5}

(2 x + 1)^{5} = 32 x^{5} + 80 x^{4} + 80 x^{3} + 40 x^{2} + 10 x + 1

= f^{^{''}} (x) (32 x^{5} + 80 x^{4} + 80 x^{3} + 40 x^{2} + 10 x + 1)

括弧を分配する

= f^{^{''}} (x) 32 x^{5} + f^{^{''}} (x) 80 x^{4} + f^{^{''}} (x) 80 x^{3} + f^{^{''}} (x) 40 x^{2} + f^{^{''}} (x) 10 x + f^{^{''}} (x) 1

= 32 x^{5} f^{^{''}} (x) + 80 x^{4} f^{^{''}} (x) + 80 x^{3} f^{^{''}} (x) + 40 x^{2} f^{^{''}} (x) + 10 x f^{^{''}} (x) + 1 f^{^{''}} (x)

乗算:

1 f^{''} (x) = f^{''} (x)

= 32 x^{5} f^{^{''}} (x) + 80 x^{4} f^{^{''}} (x) + 80 x^{3} f^{^{''}} (x) + 40 x^{2} f^{^{''}} (x) + 10 x f^{^{''}} (x) + f^{^{''}} (x)