Derivation $f(tx) = t^{\mu}f(x) \text{ } \forall x \in \mathbb{R}^n \text{\ {0}} \text{ } \forall t \in \mathbb{R}^+$

Question

Let $\mu \in \mathbb{R}$ be a real number and $f:\mathbb{R}^n$\ {$0$} $\to \mathbb{R}$ a function that is positive homegenous with degree $\mu$, which means:

$$f(tx) = t^{\mu}f(x) \text{ } \forall x \in \mathbb{R}^n \text{\ {0}} \text{ } \forall t \in \mathbb{R}^+$$

Furthermore, $f$ is totally differentiable in $\mathbb{R}^n$ \ {$0$}.

How can I show that for a fix $x \in \mathbb{R^n}$\ {$0$} the function $\mathbb{R^+} \ni t \to f(tx) \in \mathbb{R}$ is totally differentiable in every $t \in \mathbb{R}^+$ and calculate the function in the position $t$ with the multidimensional chain rule?

I know that the chain rule is

But I struggle with the function.

Answer 1

If $x \ne 0$ is fixed , then put $a:=f(x)$ and $g_x(t):=f(tx)$.

We have that $g_x(t)=t^{\mu}f(x)=a t^{\mu}$.

Can you proceed ?