Let $T$ a continuous non-negative random variable that represents a failure time. Taking $S(t)=P(T\geq t)$ and $f(t)$ the probability density function of $T$. Show that $$\lambda(t)=\frac{f(t)}{S(t)}=-\frac{d}{dt} \log S(t)$$ where $\lambda(t)$ represents a failure rate.
I had some additional information $$\lambda(t)=\frac{P(T\geq t)-P(T\geq t+\Delta t)}{\Delta t \, S(t)}$$ and
$$\lambda(t)=\lim_{\Delta t\rightarrow 0}\frac{P(t\leq T<t+\Delta t\mid T\leq t)}{\Delta t}$$
I tried to do some manipulation with that, but I can't get nowhere.
Let the distribution function of $T$ be $F(t)$, $t\in \mathbb{R}$, whose density is $f(t)$. We then have the following:
\begin{align} \lambda(t)&=\lim\limits_{h\rightarrow 0}\left(\frac{F(t+h)-F(t)}{h}\right)\frac{1}{S(t)}\\ &=\lim\limits_{h\rightarrow 0}\left(\frac{P(T\leq t+h)-P(T\leq t)}{h}\right)\frac{1}{S(t)}\\ &=\lim\limits_{h\rightarrow 0}\left(\frac{1-P(T> t+h)-P(T\leq t)}{h}\right)\frac{1}{S(t)}\\ &=\lim\limits_{h\rightarrow 0}\left(\frac{P(T>t)-P(T> t+h)}{h}\right)\frac{1}{S(t)}\\ &\stackrel{(a)}{=}\lim\limits_{h\rightarrow 0}\left(\frac{P(T\geq t)-P(T\geq t+h)}{h}\right)\frac{1}{S(t)}\\ &=\lim\limits_{h\rightarrow 0}\left(\frac{S(t)-S(t+h)}{h}\right)\frac{1}{S(t)}\\ &=-\lim\limits_{h\rightarrow 0}\left(\frac{S(t+h)-S(t)}{h}\right)\frac{1}{S(t)}\\ &=-\frac{1}{S(t)}\frac{d}{dt}S(t)=-\frac{d}{dt}\log(S(t)), \end{align} where $(a)$ above follows from the fact that $T$ is a continuous random variable, and hence $P(T> t)=P(T\geq t)$ for any $t\in \mathbb{R}$.