Let $x\mapsto f(x)$ be a function of one variable satisfying the condition $f(\alpha x) = \varphi(\alpha)f(x)$, where $f$ and $\varphi$ are differentiable functions. Show that $\varphi(\alpha)=\alpha^d $.
This exercise is form Zorich Mathematical Analysis I Chapter 8.3.4
Hint: $\varphi(\alpha \beta) = \varphi(\alpha) \varphi(\beta)$.