For example, if you have a tangent unit vector
T(t) = $\langle $$\sqrt{2}\over e^{7t}+e^{-7t}$, $e^{7t}\over e^{7t}+e^{-7t}$, $-e^{-7t}\over e^{7t}+e^{-7t}$$\rangle$
and you want to find its derivative, could you factor out $1\over e^{7t}+e^{-7t}$ to get
T(t) = $1\over e^{7t}+e^{-7t}$$\langle $$\sqrt{2}$, $e^{7t}$, $-e^{-7t}$$\rangle$
and then take the derivative of the interior term by term and the factor to get
T'(t) = $1\over 7e^{7t}-7e^{-7t}$$\langle $$0$, $7e^{7t}$, $7e^{-7t}$$\rangle$?
Any advice as to take a derivative of a vector such as this in an easier way would be welcome.
...or is brute forcing a derivative of these kinds of vectors simply inevitable?
Yes, you can pull out the factor. But when you do, because the factor depends on $t$, you'll have to use the (usual) product rule to compute the derivative:
$$\frac{d}{dt} [f(t) \cdot \vec{v}(t)] = \frac{d}{dt}f(t) \cdot \vec{v}(t) + f(t) \frac{d}{dt}\vec{v}(t)$$
In this case,
$$\frac{d}{dt}\left[\frac{1}{e^{7t} + e^{-7t}} \left\langle \sqrt{2}, e^{7t}, -e^{-7t}\right\rangle\right] =\\ \frac{d}{dt}\left[\frac{1}{e^{7t} + e^{-7t}}\right] \left\langle \sqrt{2}, e^{7t}, -e^{-7t}\right\rangle + \frac{1}{e^{7t} + e^{-7t}} \frac{d}{dt}\left\langle \sqrt{2}, e^{7t}, -e^{-7t}\right\rangle $$
$$= \frac{1}{7e^{7t}-7e^{-7t}}\left\langle \sqrt{2}, e^{7t}, -e^{-7t}\right\rangle + \frac{1}{e^{7t} + e^{-7t}} \left\langle 0, 7 e^{7t}, 7 e^{-7t}\right\rangle$$