Let \(\textbf{u}(t)\) be defined as follows:
$$\textbf{u}(t) = h(t)\textbf{i} + k(t)\textbf{j} $$
The derivative of \(f(t)\textbf{u}(t)\) is:
$$\begin{align} \frac{d}{dt} f(t) \textbf{u}(t) &= \frac{d}{dt} [f(t)h(t)\textbf{i} + f(t)k(t)\textbf{j}] \\ &= \frac{d}{dt} f(t)[h(t)\textbf{i} + k(t)\textbf{j}] \end{align}$$
According to the sum property of derivatives:
$$ \frac{d}{dt} f(t) \textbf{u}(t) = \frac{d}{dt} [f(t)h(t)]\textbf{i} + \frac{d}{dt} [f(t)k(t)]\textbf{j} $$
According to the product rule of derivatives:
$$ \frac{d}{dt} f(t) \textbf{u}(t) = \frac{d}{dt} [f'(t)h(t)+f(t)h'(t)]\textbf{i} + \frac{d}{dt} [f'(t)k(t)+f(t)k'(t)]\textbf{j} $$
Rearranging:
$$\begin{align} \frac{d}{dt} f(t) \textbf{u}(t) &= f'(t)h(t)\textbf{i}+f(t)h'(t)\textbf{i} + f'(t)k(t) \textbf{j}+f(t)k'(t)\textbf{j} \\ &= f'(t)[h(t)\textbf{i} + k(t) \textbf{j}] + f(t)[ h'(t)\textbf{i}+k'(t)\textbf{j}]\end{align}$$
This means
$$\begin{align} \frac{d}{dt} f(t) \textbf{u}(t) &= f'(t)h(t)\textbf{i}+f(t)h'(t)\textbf{i} + f'(t)k(t) \textbf{j}+f(t)k'(t)\textbf{j} \\ &= f'(t)\textbf{u}(t)+ f(t)\textbf{u}'(t)\end{align}$$