Variation of $_n d_x = l_x - l_x+n$ (Actuarial notation used) I am having trouble deriving this equation 

I understand that
$l_x=l_0 s(x)$ $s(x) = e^{-\int_0^x \mu (u)du}$ and from there I derived $l_{x+n} = l_xe^{-\int_x^{x+n}\mu(u)du}$.
I tried playing around with  $l_x - l_{x+n} = l_x\left(1-\frac{s(x+n)}{s(x)}\right) = l_x*_nq_x$
where $_nq_x$ is actuarial notation for $Pr[n+x<X \quad|\quad x<X] $.
I feel like I am missing something simple here . . . :(
Can someone help me out?
 A: Let $l(x)\equiv l_{x}$. Note that $l(x)$ is just $l(0)s(x)$. So $$\begin{align*}l(x)-l(x+n) = \int_{x+n}^{x} l'(y)\, dy = - \int_{x}^{x+n} l'(y)\, dy \tag{$\star$}\end{align*}$$ by the Fundamental Theorem of Calculus. Now, we have
$$\begin{align*}
l(y) &= l(0)s(y) \\
\Rightarrow l'(y) &= l(0) \color{blue}{s'(y)} \\
\Rightarrow l'(y) &= l(0)\times (\color{blue}{-\mu(y) s(y)}). \tag{$\star\star$}
\end{align*}$$
[The last equality is because $$s(y) = \exp\left(-\int_{0}^{y}\mu(t)\, dt\right),$$ so $$\begin{align*}s'(y) &= \exp\left(-\int_{0}^{y}\mu(t)\, dt\right) \times \frac{d}{dy}\left(-\int_{0}^{y}\mu(t)\, dt\right)\\ &= \exp\left(-\int_{0}^{y}\mu(t)\, dt\right) \times (-\mu(y))\\ &= -s(y)\mu(y),\end{align*}$$ using the Chain Rule and the Fundamental Theorem of Calculus.]
Substituting the result of $(\star\star)$ into $(\star)$ gives
$$
\begin{align*}
l(x)-l(x+n) &= -\int_{x}^{x+n} l(0)\times (-\mu(y)s(y))\, dy \\
&= \int_{x}^{x+n} l(0)s(y)\mu(y)\, dy\\
&= \int_{x}^{x+n} l(y)\mu(y)\, dy \tag{since $l(y) = l(0)s(y)$},
\end{align*}
$$
as required.
