Derivation of expressoin for hazard function (another aspect)

Question

I'm now learning the very basics of survival analysis.

Firstly, the probability function is introduced:

$$f(t)=\frac{dF(t)}{dt}=-\frac{dS(t)}{dt}$$

Where $S(t)=1-F(t)$.

We are then introduced to the hazard function:

$$h(t)=\frac{f(t)}{S(t)}$$

Subsequently, we are told that combining the previous two expressions yield:

$$h(t)=-\frac{d}{dt}log S(t)$$

How do I justify this last step, shouldn't it be something like:

$$h(t)=\frac{-\frac{dS(t)}{dt}}{S(t)}$$

??

Answer 1

Simply evaluate the derivative to obtain: $$h(t) = -\frac{d}{dt}\log(S(t)) = -\frac{1}{S(t)}\frac{dS}{dt} = \frac{f(t)}{S(t)}$$

Answer 2

From your first equation

$$ \frac{{\rm d}S}{{\rm d}t} = -\frac{{\rm d}F}{{\rm d}t} = -f(t) \tag{1} $$

So that

$$ h(t) = -\frac{{\rm d}}{{\rm d}t} \log S(t) = -\frac{1}{S(t)}\frac{{\rm d} S}{{\rm d} t} = -\frac{1}{S(t)}[-f(t)] $$