Proof of relationship between hazard rate, probability density, survival function I am reading a bit on survival analyses and most textbooks state that
$h(t)= \lim_{ \Delta t \rightarrow 0}  \frac{P(t < T \leq   t+\Delta t |T \geq t  )}{ \Delta t} =\frac{f(t)}{1-F(t)} (1)$
where $h(t)$ is the hazard rate, 
$f(t)=\lim_{\Delta t \rightarrow 0}  \frac{P(t < T \leq   t+\Delta t)}{ \Delta t}(2)$ the density function, 
$F(t)=Pr(T<t) (3)$ and 
$S(t)=Pr(T>t)=1-F(t) (4)$
Also they state that
$S(t)=  e^{- \int_0^t h(s)ds } (5)$
Most textbooks (at least those I have) do not provide proof for either (1) or (5). I think I managed to get through (1) as follows
$h(t)= \lim_{ \Delta t \rightarrow 0}  \frac{P(t < T \leq   t+\Delta t |T \geq t  )}{ \Delta t}=$
$\lim_{ \Delta t \rightarrow 0}  \frac{P(T \geq t |t < T \leq   t+\Delta t  ) P(t < T \leq   t+\Delta t)}{    P(T \geq t)\Delta t}$ which because of (2) and (4) becomes 
$\lim_{ \Delta t \rightarrow 0}  \frac{P(T \geq t |t < T \leq   t+\Delta t  )f(t)}{S(t)\Delta t}$
but $P(T \geq t |t < T \leq   t+\Delta t  )=1$ therefore $h(t)=\frac{f(t)}{1-F(t)}$
How does one prove (5)?
 A: $$h(t) = \frac{f(t)}{S(t)}\ $$
$$ = \frac{f(t)}{1-F(t)}$$
$$= \frac{f(t)}{1- \int^t_0{f(s) ds}}$$
Integrate both sides:
$$\int^t_0 h(s) ds = \int^t_0 \frac{f(s)}{1- \int^t_0{f(s)ds}}ds $$
$$= -\ln [1- \int^t_0{f(s)ds}]^t_0+ c $$
$$1- \int^t_0{f(s)ds} = \exp [-\int^t_0 h(s) ds]$$
Differentiate both sides:
 $$-f(t) = -h(t) \exp[-\int^t_0 h(s) ds]$$
$$f(t) = h(t) \exp[-\int^t_0 h(s) ds]$$
Since $$h(t) = \frac{f(t)}{S(t)}$$
$$S(t) = \frac{f(t)}{h(t)}$$
Replace $f(t)$ by $h(t) \exp[-\int^t_0 h(s) ds]$ , 
$$S(t) = \frac{h(t) \exp[-\int^t_0 h(s) ds]}{h(t)}$$
Therefore,
$$S(t) = \exp[-\int^t_0 h(s) ds]$$
A: We prove the following equation:
$$
S(t)=\exp\{-\int_{0}^{t}h(u)du\}
$$
proof:
We first prove
$$
f(t)=-\frac{dS(t)}{dt}
$$
proof:
$$
f(t)=\frac{dF(t)}{dt}=\frac{dP(T<t)}{dt}=\frac{d(1-S(t))}{dt}=-\frac{dS(t)}{dt}\ \blacksquare
$$
And we know
$$
h(t)=\frac{f(t)}{S(t)}
$$
Substitute $f(t)$ into $h(t)$ we get
$$
h(t)=\frac{-\frac{dS(t)}{dt}}{S(t)}
$$
then continue our main proof. By integrate the both side of the above equation, we have
$$
\int_0^th(u)du=\int_0^t\frac{-\frac{dS(t)}{dt}}{S(t)}dt=\int_0^t-S(t)^{-1}dS(t)\\
=-[\log S(t)-\log S(0)]=-\log S(t)
$$
Then we get the result
$$
S(t)=\exp\{-\int_0^th(u)du\}\ \blacksquare
$$
A: The derivative of $S$ is
$$
\frac{\mathrm{d}S(t)}{\mathrm{dt}} = \frac{\mathrm{d}(1 - F(t))}{\mathrm{dt}} = - \frac{\mathrm{d}F(t)}{\mathrm{dt}} = -f(t)
$$
Therefore, as mentioned by @StéphaneLaurent, we have
$$
-\frac{\mathrm{d}\log(S(t))}{\mathrm{dt}} = \cfrac{-\frac{\mathrm{d}S(t)}{\mathrm{dt}}}{S(t)} = \frac{f(t)}{S(t)} = h(t) 
$$
where the last equality follows from (1). 
Taking the integral both sides of the previous relation, we obtain
$$
-\log(S(t)) = \int_0^t h(s) \, \mathrm{d}s
$$
so that
$$
S(t) = \exp \left\{- \int_0^t h(s) \, \mathrm{d}s\right\}
$$
This is your equation (5). The integral part in the exponential is the integrated hazard, also called cumulative hazard $H(t)$ [so that $S(t) = \exp(-H(t))$].
