Prove a condition for a vector valued function to be differentiable

If $F$ is a vector valued function, prove that $F$ is differentiable on the open interval $I$ if and only if for all $t \in I$ we have

$F'(t) = \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)).$

Proof. First, assume $F$ is differentiable on $I$ . Then, by the definition of differentiability for vector valued functions we have

$F'(t) &= (f_1'(t), \ldots, f_n'(t))$

and each of the $f_i'(t)$ exists. Therefore by the usual definition of the derivative we have

$\begin{align*} F'(t) &= \left( \lim_{h \to 0} \frac{f_1 (t+h) - f_1(t)}{h}, \ldots, \lim_{h \to 0} \frac{f_n (t+h) - f_n (t)}{h} \right) \\[9pt] &= \lim_{h \to 0} \frac{1}{h} \left( (f_1 (t+h), \ldots, f_n (t+h)) - (f_1 (t), \ldots, f_n (t)) \right) \\[9pt] &= \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)). \end{align*}$