Define $f=\underset {i\in \{1,\ldots,n\}}\max f_i$ where all $f_i: \mathbb R ^n \rightarrow \mathbb R$ are differentiable. Let $0\ne d\in \mathbb R ^n$. I need to show that $f'(x;d)=\underset {i\in J}\max f_i'(x;d)$ where $J$ is the set $\{i\mid f(x)=f_i (x)\}$ and $f'(x;d)$ is the directional derivative of $f$ at the point $x$ in the direction $d$.
Trying to prove this using the definition of directional derivative, I find that I need to show that $\lim _{h \rightarrow 0}f(x+hd)=\lim _{h \rightarrow 0}f_i(x)$ where $f(x)=f_i (x)$. But how can I show that these two limits are equal rigorously? And how can I conclude the equality $f'(x;d)=\underset {i\in J}\max f_i'(x;d)$?
$(*)$ One important thing you need to know for this problem is that if real-valued differentiable functions agree on an interval then they give the same directional derivative.
And so the maximum is attained using some point $x_*$ in this interval, moreover, by intermediate value theorem $f_j(x)>f_i(x)$ then $f_j(t)> f_i(t)$ where $t \in (x- \delta, x + \delta)$.
Hence if $f(x) = \textbf{max} \{f_1(x),...,f_n(x)\}$ then choosing $f_i(x) = f(x)$ we know that $f_i(x)>f_{ j} (x)$ for all $j \not = i$ and for some interval about $x$.
Thus, $D_{\vec{v}} f (x) \cdot t \approx f(x+t\vec{v}) - f(x) = f_i(x+t\vec{v}) - f_i(x) \Rightarrow D_{\vec{v}}f(x) = D_{\vec{v}}f_i(x) $.
Using $(*)$ we know that if any $f_j$ agree with $f_i$ on any neighborhood of $f_i$'s domain then $D_{\vec{v}}f_j(x) = D_{\vec{v}}f_i(x)$.
So in the way in which I would go about this, I would just have to prove $(*)$, right? Well, suppose $f(x) = g(x)$ where $f,g$ are differentiable and real-valued. If we define $h = f-g$ then $h' = 0$ and so $h$ is ? Hence, $f,g$ differ by a ? From the relation you get in answering the preceding question you'll see the equivalent derivative result immediately.