Define the directional derivative of a function $\textbf{f}$ at $\textbf{c}$ in the direction $\textbf{u}$ by $$\textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u}) = \lim_{h \rightarrow 0} \frac{\textbf{f}(\textbf{c}+h\textbf{u}) - f(\textbf{c})}{h},$$ whenever the limit on the right exists.
Let $\textbf{f}$ and $\textbf{g}$ be functions with values in $\mathbb{\textbf{R}}^m$ such that the directional derivatives $\textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u})$ and $\textbf{g}'(\textbf{c};\textbf{u})$ exist.
Prove that the sum $\textbf{f+g}$ and dot product $\textbf{f}$ $\bullet$ $\textbf{g}$ have directional derivatives given by $$(\textbf{f+g})'(\textbf{c};\textbf{u}) = \textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u}) + \textbf{g}'(\textbf{c};\textbf{u})$$ and $$(\textbf{f} \bullet \textbf{g})'(\textbf{c};\textbf{u)} = \textbf{f}(\textbf{c}) \bullet \textbf{g}'(\textbf{c};\textbf{u}) + \textbf{g}(\textbf{c}) \bullet \textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u}).$$
$\underline{\textbf{What I've Tried}}$ I've figured out how to prove the sum part already but I'm having trouble with the dot product part. I tried to do it backwards
$f(c) \bullet g'(c;u) + g(c) \bullet f'(c;u)$
$= f(c) \bullet \lim_{h \rightarrow 0} \frac{g(c+hu)-g(c)}{h} + g(c) \bullet \lim_{h \rightarrow 0} \frac{f(c+hu)-f(c)}{h}$
$= \lim_{h \rightarrow 0} f(c) \bullet \frac{g(c+hu)-g(c)}{h} + \lim_{h \rightarrow 0} g(c) \bullet \frac{f(c+hu)-f(c)}{h}$
$=\lim_{h \rightarrow 0} \frac{f(c) \bullet g(c+hu) - f(c) \bullet g(c) + g(c) \bullet f(c+hu) - g(c)\bullet f(c)}{h}$
$\textbf{But I'm not sure how that will arrive to}$:
$= \lim_{h \rightarrow 0} \frac{f(c+hu)\bullet g(c+hu) - f(c) \bullet g(c)}{h}$
$= \lim_{h \rightarrow 0} \frac{(f \bullet g)(c+hu) - (f \bullet g)(c)}{h}$
$= (f \bullet g)'(c;u)$
Hint: when $h\to0$, $$\begin{align}(f\cdot g)(c;u)&=\frac{f(c+hu)\cdot g(c+hu)-f(c)\cdot g(c)}{h}\\ &=\frac{f(c+hu)\cdot g(c+hu)-f(c+hu)\cdot g(c)+f(c+hu)\cdot g(c)-f(c)\cdot g(c)}h\\ &=f(c+hu)\cdot g'(c;u)+f'(c;u)\cdot g(c)\\ &=f(c)\cdot g'(c;u)+f'(c;u)\cdot g(c)\end{align}$$