I am can not figure out how to solve this prblem:
Suppose that $f:A\subseteq\mathbb{R}^m\rightarrow\mathbb{R}^n$, where $A$ is a neighborhood of $a\in\mathbb{R}^m$. Suppose that all directional derivatives of $f$ exist at $a$, and suppose that $f^\prime(a;u+v)=f^\prime(a;u)+f^\prime(a;v)$ for all $u,v\neq0$ such that $u+v\neq 0$. Let $J$ denote the Jacobian of $f$ at $a$, so that $J$ is the $n\times m$ matrix whose $(i,j)$-entry is $D_jf_i(a)$. Show that $Ju=f^\prime(a;u)$ for all nonzero $u\in\mathbb{R}^m$.
I did the matrix multiplication, but I can not get $D_j f_k(a) \cdot u_k=f'_k(a,u).$
Any hints would be appreciated.
The Jacobian matrix of $f$ at $a$ must be satisfy that $$\lim_{h \to \mathbf 0} \frac{f(a+h) - f(a) - Jh}{\|h\|} = \mathbf 0,$$ in particular: $$\mathbf 0 = \lim_{t \to 0} \frac{f(a+tu) - f(a) - J(tu)}{\|tu\|} = \lim_{t \to 0} \bigg(\frac{1}{\|u\|} \frac{f(a+tu) - f(a)}{|t|} - \frac{t}{|t| \|u\|} Ju \bigg)$$ and multiplying the above by $\|u\|$ we have: $$\mathbf 0 = \lim_{t \to 0} \bigg(\frac{f(a+tu) - f(a)}{|t|} - \frac{t}{|t|} Ju \bigg).$$ Now, note that $$\mathbf 0 = \lim_{t \to 0^+} \bigg(\frac{f(a+tu) - f(a)}{t} - Ju \bigg) = f'(a;u)-Ju$$ and that $$\mathbf 0 = \lim_{t \to 0^-} \bigg(-\frac{f(a+tu) - f(a)}{t} + Ju \bigg) = -f'(a;u)+Ju.$$