Thank you for your reply. My follow up question is
How did I know that I can't use "T" as the variable given that in the first equation I am taking a derivative with respect to "T" on the right hand side?
But I think I can see it if I work backwards:
$\displaystyle \log{P(t,T)} = -\int_{t}^{T}f(t,u)du$
$\displaystyle \log{P(t,T)} = -g(t,t) + g(t,T) + C$
where $\displaystyle f(t,u) = \frac{\partial g(t,u)}{\partial u}$.
I know that at $\displaystyle T=t, \log{P(t,t)} = 0 $ so $\displaystyle C=0$
Next take the partial derivative of both sides with respect to "T" such that
$\displaystyle \frac{\partial \log{P(t,T)}}{\partial T} = -\frac{\partial g(t,t)}{\partial T} + \frac{\partial g(t,T)}{\partial T}$
$\displaystyle \frac{\partial \log{P(t,T)}}{\partial T} = f(t,T)$
by our earlier definition of $\displaystyle f(t,u) = \frac{\partial g(t,u)}{\partial u}$.
Comments welcome. Thanks again