f-x-x-2n-Determinate-f-n-x-

Question Number 166813 by mathocean1 last updated on 28/Feb/22

f (x) = x^{2 n}

D e t e r m i n a t e f^{(n)} (x)

Answered by TheSupreme last updated on 28/Feb/22

f^{'} (x) = α x^{α - 1}

f^{(k)} (x) = \frac{2 n!}{(2 n - k)!} x^{2 n - k}

Answered by Mathspace last updated on 28/Feb/22

f^{(1)} (x) = 2 {n x}^{2 n - 1}

f^{(2)} (x) = 2 n (2 n - 1) x^{2 n - 2}

b y r e c u r r e n c e f^{(p)} (x) = 2 n (2 n - 1) \dots (2 n - p + 1) x^{2 n - p}

p = n \Rightarrow f^{(n)} (x) = 2 n (2 n - 1) \dots . (n + 1) x^{n}

= \frac{2 n (2 n - 1) \dots (n + 1) n!}{n!} x^{n}

= \frac{(2 n)!}{n!} x^{n}