If a zero coupon bond price at time $t$, with maturity $~T~~~ (t<T)~$, is denoted by
$$B(t;T) = B(T;T) ~e^{-\int_{t}^{T} r(s) ds}$$
where $r(t)$ is a known interest rate.
How does this transform to $$r(T) = - \frac{1}{B(t;T)} \frac{\partial B(t;T)}{\partial T}$$
I know that $~B(T;T)=1~$ and we can rearrange, but I don't understand how to obtain partial differentials from integrals?
$$B(t,T) = B(T,T) ~e^{-\int_{t}^{T} r(s) ds}=e^{-\int_{t}^{T} r(s) ds}\Leftrightarrow$$
$$ -\log B(t,T)= \int_{t}^{T} r(s) ds \text{ } \text{ } \text{ }(1) $$ Apply chain rule to LHS to find: $$\frac{\partial}{\partial T}(-\log B(t,T))=-\frac{1}{B(t,T)}B_T(t,T)$$ with $B_T(t,T) = \frac{\partial B(t;T)}{\partial T}$
By the Second Fundamental Theorem of Calculus RHS can be expressed as: $$\frac{\partial}{\partial T}\int_{t}^{T} r(s) ds=r(T)$$
Now Equation (1) can be expressed as $$- \frac{1}{B(t,T)} B_T(t,T) = r(T) $$ just as you wanted to show