$ U ⊂ R^m$ is open and we have $f : U → R^n$. Prove if $df_a(u)$ exists, then $df_a(bu)$ exists and is equal to $b*df_a(u)$ for any $b \in R$ that is not 0.
I assumed $df_a(u)$ is a directional vector in the direction of a, and I used chain-rule to derive the result but I am not sure if it applies to directional vectors.
This was actually a quiz question that I couldn't get and hope to understand before my midterm. Thanks for the help.
It appears that $df_a(u)$ denotes the directional derivative of $f$ at a point $a$ in the direction of $u$ (not the direction of $a$!). The result in question then indeed follows from the chain rule (for ordinary functions of a single variable; no need to worry about a chain rule for directional derivatives): $df_a(u)$ is by definition $g'(0)$ where $g$ is defined by $g(t)=f(a+tu)$, and $df_a(bu)$ is by definition $h'(0)$ where $h(t)=f(a+tbu)$. But $h(t)=g(bt)$ so by the chain rule $h'(0)=bg'(0)$.
If you are not comfortable using the chain rule in this way, this fact is also easy to verify directly from the definition of the directional derivative as a limit: you wish to show $$\lim_{t\to 0}\frac{f(a+tbu)-f(a)}{t}=b\cdot\lim_{t\to 0}\frac{f(a+tu)-f(a)}{t}$$ assuming the limit on the right exists. To do so, just substitute $t'=tb$ on the left side and observe that a limit as $t\to 0$ is equivalent to a limit as $t'\to 0$.