5 votes 5 votes $n$-th derivative of $x^n$ is $nx^{n-1}$ $n^n.n!$ $nx^n!$ $n!$ Calculus isro2011 calculus differentiation + – go_editor asked Jun 23, 2016 • edited Dec 8, 2022 by Lakshman Bhaiya go_editor 2.0k views answer comment Share Follow See 1 comment See all 1 1 comment reply vijaycs commented Jun 23, 2016 reply Follow Share Ans - D. n! 0 votes 0 votes Please log in or register to add a comment.
Best answer 8 votes 8 votes f(x) = xⁿ f'(x) = n x(n-1) f''(x) = n(n-1) x(n-2) f'''(x) = n(n-1)(n-2) x(n-3) fⁿ(x) = n! x(n-n) , and since n-n = 0, x0 =1, so fⁿ(x) = n! Hence,Option(D)n! is the correct choice. LeenSharma answered Jun 23, 2016 • selected Jun 23, 2016 by srestha LeenSharma comment Share Follow See all 0 reply Please log in or register to add a comment.
