I need to find the unit normal vector given:

Question

r(t)= ti + (6/t) j where t= 3

I know I need to find the unit tangent vector first, which is r'(t) divided by the magnitude of r'(t).
r(t) can be rewritten as ti+ 6t^-1
r'(t) = i- 6t^-2 = i- [6/ (t^2)] j
For the magnitude of r'(t) I got the square root of (t^4 +36) all divided by t^2.
For T(t), I got (it^2 -6j)/ (sq. rt. t^4 +36)

Is this right? If so, I don't know how to go about finding T'(t) or the magnitude of T'(t). Please help! Also, where does the t= 3 come in?