n. [ f(x)] n-1.f'(x) = n.f'(nx)
[f(x)]n.[f(x)]-1 .f'(x) = f'(nx)
[f(x)]n .f'(x) = f'(nx).f(x) -------1
now
from ques
[f(x)]n = f(nx)
therefore from ----1
f(nx).f'(x) = f'(nx).f(x)
1. A function f(x) is defined so that for all real x {f(x)}n = f(nx). Prove that
f(x).f'(nx) = f'(x).f(nx).
n. [ f(x)] n-1.f'(x) = n.f'(nx)
[f(x)]n.[f(x)]-1 .f'(x) = f'(nx)
[f(x)]n .f'(x) = f'(nx).f(x) -------1
now
from ques
[f(x)]n = f(nx)
therefore from ----1
f(nx).f'(x) = f'(nx).f(x)