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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*}

Conversely, assume that

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

Then, the limit

    \[ \lim_{h \to 0} \frac{1}{h} (f_i (t+h) - f_i(t)) \]

exists for each i = 1, \ldots, n. Thus, f_i'(t) exists for each i. Therefore, F'(t) exists. \qquad \blacksquare