derivative f(x)=n(n^2+3n-4)^7

解

導関数

f (x) = n (n^{2} + 3 n - 4)^{7}

(n^{2} + 3 n - 4)^{7} + 7 n (n^{2} + 3 n - 4)^{6} (2 n + 3)

解答ステップ

\frac{d}{d n} (n (n^{2} + 3 n - 4)^{7})

積の法則を適用:

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

f = n, g = (n^{2} + 3 n - 4)^{7}

= \frac{d n}{d n} (n^{2} + 3 n - 4)^{7} + \frac{d}{d n} ((n^{2} + 3 n - 4)^{7}) n

\frac{d n}{d n} = 1

\frac{d}{d n} ((n^{2} + 3 n - 4)^{7}) = 7 (n^{2} + 3 n - 4)^{6} (2 n + 3)

= 1 \cdot (n^{2} + 3 n - 4)^{7} + 7 (n^{2} + 3 n - 4)^{6} (2 n + 3) n

簡素化

= (n^{2} + 3 n - 4)^{7} + 7 n (n^{2} + 3 n - 4)^{6} (2 n + 3)