Reliability function, proving exponential distribution We are given $R(t)$ = $P(X>t)$ for all $x > 0$ and
$$R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0$$
The random variable $X$ also satisfies the memoryless property:
$$P(X>s+t|X>t) = P(X>s)\text{ for }s>0\text{ and }t>0$$
Let $R'(0) = - \lambda\ where \  \lambda\ ,$  is some positive constant. I need to show that X must be exponentially distributed.
Given that $\dfrac{R(t + h) - R(t)}{h}$ = $R(t)\left[\dfrac{R(h) - 1}{h}\right]$ 
Show that by letting $\lim\limits_{h\to \infty}$ $\dfrac{dR(t)}{dt} = -\lambda$$R(t)$  (I think we should use Hopital's rule here I am not sure by differentiating $\left[\dfrac{R(h) - 1}{h}\right]$ and letting $h$ tend to $0$, we will get $-\lambda$ for this part but I got stuck afterwards).
Also argue that $X$ is an exponential random variable with rate parameter $\lambda$  by solving the differential equation above respecting the conditions:
$$R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0$$
 A: Below is the standard argument, which is undoubtedly done in essentially the same way in your textbook.  After a small amount of probability, the rest is just calculus. We are told that 
$$P(X>s+t\,|\,X>t) = P(X>s). \qquad\qquad(\ast)$$
Let $A$ be the event $X>s+t$, and let $B$ be the event $X>t$. We know that 
$$P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}.$$
In this case, $P(A\cap B)=P(A)$. So
$$P(X>s+t\,|\,X>t) = \frac{P(X>s+t)}{P(X>t)}.$$ 
Using $(\ast)$, we conclude that 
$$\frac{P(X>s+t)}{P(X>t)}=P(X>s).$$
This may be more compactly rewritten as 
$$R(s+t)=R(s)R(t).$$
Precisely this equation was given in a comment by Dilip Sarwate.
To put you on more familiar ground, we replace $s$ by $x$, and $t$ with $h$.  So we have reached the equation
$$R(x+h)=R(x)R(h).\qquad\qquad(\ast\ast)$$
Subtract $R(x)$ from both sides, and then divide by $h$. We arrive at
$$\frac{R(x+h)-R(x)}{h}=R(x)\frac{R(h)-1}{h}.$$
Let $h$ approach $0$. As $h$ approaches $0$, the right-hand side, by definition, approaches $R'(0)$, which we were told is $-\lambda$. So the left-hand side has a limit, which by definition is $R'(x)$.
(By only considering positive $h$ only, we are being a little dishonest. Don't worry about it too much.)
So after some calculation, we have reached the differential equation
$$R'(x)=-\lambda R(x), \quad\text{or if you prefer}\quad\frac{dR}{dx}=-\lambda R.$$
This is the familiar differential equation for exponential decay.  The general solution is 
$$R(x)=R(0)e^{-\lambda x}.$$
We were told that $R(0)=1$. It follows that $R(x)=e^{-\lambda x}$.
So the cumulative distribution function $F_X(x)$ of the random variable $X$ is $1-e^{-\lambda x}$ (for $x>0$).  Differentiate with respect to $x$. We conclude that the probability density function $f_X(x)$ of $X$ is $\lambda e^{-\lambda x}$ (for $x>0$).
