let $ K\subseteq M \subseteq L$. If the extensions $M \subseteq L$ and $K \subseteq M$ are separable, then $K\subseteq L$ is separable. Let L be a finite extension of K and let $ K\subseteq M \subseteq L$. Show that if the extensions $M \subseteq L$ and $K \subseteq M$ are separable, then $K \subseteq L$ is separable.
Here is what we know:
$1. M \subseteq$ L separable: 
$\forall l \in L$ the minimal polynomial $m_{l,M}(x)$ is separable in $M$.
$2. K \subseteq$ M separable: 
$\forall n \in L$ the minimal polynomial $m_{n,K}(x)$ is separable in $K$.
What we want: 
$ K \subseteq$ L separable: 
$\forall l \in L$ the minimal polynomial $m_{l,K}(x)$ is separable in $K$.
If  $l\in M$, then it works by hypothesis. However, if $l \in L/M,$ then I don't know how to do it.
 A: All you have to do is take $\theta\in L$ and show that it is separable in $K$. Take the irreducible polynomial of $\theta$ over $M$, $\mathrm{Irr}(\theta,M)(X)\in M[X]$ and denote its coefficients by $a_0,\dots,a_n$. Since $M|K$ is a separable extension, the coefficients are separable over $K$, so we have that the extension $K(a_0,\dots,a_n)|K$  is separable.
On the other hand, the fact that $\theta$ is separable over $K(a_0,\dots,a_n)$ implies that the irreducible polynomial $\mathrm{Irr}(\theta,M)(X)$ does not have roots with multiplicity greater than one. Also, $\mathrm{Irr}(\theta,M)(X)\in K(a_0,\dots,a_n)[X]$. 
Hence, 
$$[K(a_0,\dots,a_n,\theta):K]_s=[K(a_0,\dots,a_n,\theta):K(a_0,\dots,a_n)]_s[K(a_0,\dots,a_n):K]_s=[K(a_0,\dots,a_n,\theta):K(a_0,\dots,a_n)][K(a_0,\dots,a_n):K]=[K(a_0,\dots,a_n,\theta):K].$$
Therefore, $\theta$ is separable over $K$, and from that follows the result.
$\textbf{Note}:$ Remember that in a finite extension of fields $K|k$, the fact that every element of $K$ is separable over $k$ is equivalent to $[K:k]_s=[K:k]$ where $[K:k]_s$ denotes the separability degree of the extension.
