How is Hubble's equations H(t) = 2/3 t^-1 derived?

[damasgate]

I know that H(t) = 1/a(t) x da/dt

I know that a(t) is proportional to t^2/3

and da/dt is proportional to a^-1/2

now H(t) should equal H(t) = a(t)^-1 x a(t)^-1/2
= a(t)^-3/2
= (t^2/3)^-3/2
= t^-1

However where is the constant 2/3 coming from? I can't find it anywhere in my calculations!

 

[BillSaltLake]

When you differentiate t^2/3, you'll get 2/3 t^-1/3, but you dropped the "2/3" when you said "da/dt is proportional to a^-1/2". I think that's what you're asking.
BTW at present, a is more like t¹.