let $\displaystyle \vec T(t) = \frac{1}{\sqrt{10+ 4t^2}} ( 3 \hat i - \hat j + 2t \hat k) find \frac {d \vec T}{dt}$
I am not sure how to do this.
Note that
$\displaystyle \frac{d}{dt} \frac{1}{\sqrt{10 + 4t^2}} = -4t (10+4t^2)^{-\frac{3}{2}}$, and
$\displaystyle \frac{d}{dt} \frac{2t}{\sqrt{10 + 4t^2}} = \frac{2 \sqrt{10 + 4t^2} - t(10 + 4t^2)^{-\frac{1}{2}}}{10 + 4t^2}$ (applying the chain and quotient rules)
$\displaystyle \frac{d \vec{T}}{dt}$ is computed by taking the derivatives of each of vector T's components, i.e.
$\displaystyle \frac{d \vec{T}}{dt} = \frac{d}{dt} \frac{3}{\sqrt{10 + 4t^2}}\vec{i} - \frac{d}{dt} \frac{1}{\sqrt{10 + 4t^2}} \vec{j} + \frac{d}{dt} \frac{2t}{\sqrt{10 + 4t^2}} \vec{j}$
Use the given derivatives and substitute.