That depends upon what you mean by "an odd function about x= a"!
If the graph of y= f(x) is symmetric about x= a, then the graph of y= f(x+ a) is symmetric about x= 0- an even function. It follows that y= f'(x+ a) is an odd function- "symmetric through the origin" and so y= f'(x) is "symmetric through (a, 0), not necessarily an "odd function".
(Note that since f'(x+a) is an odd function, f'(a)= 0.)