Let f be a differentiable function satisfying $[f(x)]^n=f(n x)$ for all $x \in R$. Then, $f'(x) f(n x)$

$f(x) f'(n x)$

We have,

$[f(x)]^n=f(n x)$ for all x

$\Rightarrow n[f(x)]^{n-1} f'(x)=n f'(n x)$

$\Rightarrow n[f(x)]^n f'(x)=n f(x) f'(n x)$            [Multiplying both sides by f(x)]

$\Rightarrow n f(n x) f'(x)=n f(x) f'(n x)$                [∵ [f(x)]ⁿ = f(nx)]

$\Rightarrow f(n x) f'(x)=f(x) f'(n x)$