Prove the Dot Product Rule: If $c_1(t)$ and $c_2(t)$ are differentiable curves in $\mathbb R^n$ then: $$\frac{\mathrm d}{\mathrm dt} [c_1(t) \cdot c_2(t)] = c′_1(t) \cdot c_2(t) + c_1(t) \cdot c′_2(t).$$
I tried using arbitrary curves, however I am stuck. Any suggestions on how to go about this proof? Thank you.
As said by @Vercassivelaunos, using components and the Product Rule for real-valued functions works well. The classic proof for the Product Rule for real-value functions can also be adapted to this situation: let $a\in dom(c_1)\cap dom(c_2)$. Then
$$ \frac{c_1\cdot c_2(t) - c_1\cdot c_2(a)}{t-a} = \frac{c_1(t)\cdot c_2(t) - c_1(a)\cdot c_2(a)}{t-a} = \frac{(c_1(t)-c_1(a))\cdot c_2(t) + c_1(a)\cdot(c_2(t) - c_2(a))}{t-a} = \frac{c_1(t)-c_1(a)}{t-a}\cdot c_2(t) + c_1(a)\cdot\frac{c_2(t) - c_2(a)}{t-a} $$
Note that the equalities are true since the dot product is bilinear. Now, since $c_1$ and $c_2$ are differentiable at $a$, we have
$$ \lim_{t\to a}\frac{c_1(t)-c_1(a)}{t-a} = c_1'(a) \text{ and } \lim_{t\to a}\frac{c_2(t)-c_2(a)}{t-a} = c_2'(a) $$
Also, $c_2$ is differentiable at $a$, so $c_2$ is continuous at $a$, therefore $\lim_{t\to a}c_2(t)=c_2(a)$. Using Limit Laws, we have
$$ \frac{c_1\cdot c_2(t) - c_1\cdot c_2(a)}{t-a} = c_1'(a)\cdot c_2(a) + c_1(a)\cdot c_2'(a) $$
Since we use only the bilinearity of the dot product and the Limit Laws (for vector functions), then we have a similar formula for the cross product.