Why are the derivatives of periodic functions periodic?

1 Answer
Oct 30, 2015

See the explanation section below.

Explanation:

#f# is periodic if and only if there is an integer #c# such that #f(x+c) = f(x)# for all #x# in #"Dom"(f)#

If #f# is periodic for some #c# as above and #f'(x)# exists, then #f'(x+c)# exists and #f'(x+c) = f'(x)#

Use the definition of derivative. (You could use the chain rule instead, but it is not necessary.)

#f'(x+c) = lim_(hrarr0)(f((x+h)+c)-f(x+c))/h# #" "# (def of derivative)

# = lim_(hrarr0)(f((x+h))-f(x))/h# #" "# (periodicity of #f#)

# = f'(x)# #" "##" "# (def of derivative)

Therefore, #f'# is also periodic.