How do you find the nth derivative of the function #f(x)=x^n#?

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Answer 1 · 2017-01-28T19:47:39

#n!#

#f(x)=x^n#
#f'(x)=nx^(n-1)#'
#f''(x)=n(n-1)x^(n-2)#
...
#f^((n))(x)=n(n-1)...3.2.1x^0#
#=n!#

Or by induction on #n# if you want a formal proof.

Answer 2 · 2017-01-28T21:24:06

Each derivative gives us a pattern.

#f'(x) = nx^(n-1)#

#f''(x) = n(n-1)x^(n-2)#

#f'''(x) = n(n-1)(n-2)x^(n-3)#

and so on until #n - k = 0# where #k# is the order of the derivative. When we finish, we get:

#f^((k))(x) = n(n-1)(n-2)cdots(n-k+1)x^(n-k)#

When we go all the way to #n = k#, then:

#color(blue)(f^((n))(x) = n(n-1)(n-2)cdots(1)cancel(x^0)^(1))#

which is a constant equaling #color(blue)(n!)#, as #n! = n(n-1)(n-2)cdots(2)(1)#, and #x^0 = 1# which doesn't affect #n!#.