Proof Derivative Survival Ce98Fd
1. **Problem statement:** Prove the equation $$\frac{d}{dt} p_x(t) = -p_x(t) \mu_{x+t}$$ where $p_x(t)$ is the probability of survival from age $x$ to $x+t$ and $\mu_{x+t}$ is the force of mortality at age $x+t$.
2. **Recall definitions:**
- The survival function $p_x(t) = P(T_x > t) = S_x(t)$ where $T_x$ is the future lifetime of a person aged $x$.
- The force of mortality $\mu_{x+t} = -\frac{d}{dt} \ln p_x(t)$.
3. **Start with the definition of force of mortality:**
$$\mu_{x+t} = -\frac{d}{dt} \ln p_x(t) = -\frac{1}{p_x(t)} \frac{d}{dt} p_x(t)$$
4. **Rearrange to isolate $\frac{d}{dt} p_x(t)$:**
$$\frac{d}{dt} p_x(t) = -p_x(t) \mu_{x+t}$$
5. **Interpretation:** This shows the instantaneous rate of change of the survival probability is proportional to the negative of the survival probability times the force of mortality.
This completes the proof.