Derivations and separability of field extensions This is written on page 4 of James E. Humphreys' Linear Algebraic Groups:

A derivation $\delta: E \rightarrow L$ ($E$ a field, $L$ an extension field of $E$), is a map which satisfies $\delta(x+y) = \delta(x)+\delta(y)$ and $\delta(xy)=x \delta(y)+ \delta(x)y$. If $F$ is a subfield of $E$, $\delta$ is called an $F$-derivation if in addition $\delta(x)=0$ for all $x \in F$ (so $\delta$ is $F$-linear).

Then the author said:

The space $\operatorname{Der}_F(E,L)$ of all $F$-derivations $E \rightarrow L$ is a vector space over $L$, whose dimension equals the transcendence degree of $E$ over $F$ if $E/F$ is separably generated. $E/F$ is separable if and only if all derivations $F \rightarrow L$ extend to derivations $E \rightarrow L$ ($L$ an extension field of $E$).

I am confused by the last statement. Let $F = F_5 = \{0,1,2,3,4 \} $, the field of $5$ elements, and $E$ the splitting field of $x^5 -2$ over $F$. For any extension field $L$ of $E$, any derivation $\delta: F \rightarrow L$ must be the zero map (because of the two conditions). So the zero map from $E$ to $L$ is the extension of $\delta$ to $E$. But, $E/F$ is obviously inseparable.
Where am I wrong? Are there any references or hints as to the proof of the statement?
Sincere thanks.

I was wrong because $F$ is perfect and the extension is trivial.
Now, please allow me to ask for some references or hints as to the proof of:

$E/F$ is separable if and only if all derivations $F\rightarrow L$ extend to derivations $E\rightarrow L$, where $L$ is an extension field of $E$.

 A: Your error is in thinking that $E/F$ is inseparable.
First, note that $x^5-2$ is not irreducible: by Fermat's Little Theorem, $a^5 = a$ for all $a\in F$, so $2$ is a root of $x^5-2$ in $F$. Thus, the splitting field is actually the splitting field of an irreducible of degree less than $p$, and so will be separable (e.g., by the derivative test).
Also, note that finite fields are perfect (a field $k$ is perfect if and only if $k$ is of characteristic $0$, or $k$ is of characteristic $p$ and $k^p=k$). If $k$ is a perfect field, then every algebraic extension of $k$ is perfect. (E.g., Corollary V.6.12 in Lang's Algebra).
All finite fields are perfect, since the Frobenius map $x\mapsto x^p$ is an automorphism ($(a+b)^p = a^p+b^p$ and $(ab)^p =a^pb^p$; and $a^p=0$ if and only if $a=0$). So every algebraic extension of a finite field is necessarily separable. 

The book you quote has three explicit references to the statement in question, yet you are asking for "references or hints"...
Lang's Algebra (Revised 3rd Edition), Chapter VIII, Section 5, page 370, has:
Let $K(x)$ be an extension of $K$, and let $D$ be a derivation of $K$. Let $f(X)$ be the irreducible polynomial satisfied by $x$ over $K$, and let $D$ be a derivation on $K$.
If $x$ is separable algebraic over $K$, let $f(X)$ be the irreducible polynomial of $x$ over $K$. Then $f'(x)\neq 0$,  and if we define 
$$u = -\frac{f^D(x)}{f'(x)},$$
where $f^D$ is the polynomial obtained from $f$ by applying $D$ to the coefficients of $f$, then for any $g(x),h(x)\in K[x]$, $h(x)\neq 0$, we define
$$\begin{align*}
D^*g(x) &= g^D(x) + g'(x)u\\
D^*(g/h) = \frac{hD^*g - gD^*h}{h^2}
\end{align*}$$
gives a derivation on $K(x)$ which is an extension of $D$.
If $x$ is transcendental over $K$, then you can extend $D$ by defining it as above with $u$ an arbitrary element of $K(x)$. 
And if $x$ is purely inseparable over $K$, so $x^p-a=0$ for some $a\in K$, then $D$ extends to $K(x)$ if and only if $Da=0$; so if $D$ is trivial on $K$, you can extend it by selecting $u$ arbitrarily again.
Then we have:

Proposition 5.2. A finitely geenrated extension $K(x_1,\ldots,x_n)$ over $K$ is separable algebraic if and only if every derivation $D$ of $K(x_1,\ldots,x_n)$ which is trivial on $K$ is trivial on $K(x_1,\ldots,x_n)$.

Proof. If $K(x_1,\ldots,x_n)$ is separable, then one can extend by applying the first case above repeatedly. Is the extension is not separable, we can decompose it into a tower of extensions between $K$ and $K(x)$ such that each step is either separable, transcendental, or purely inseparable. At least one step must be of one of the latter two types. The the uppermost such step, and use it to construct a derivation which is trivial on the base but not on the top to establish the converse.
This gives you how to extend the derivation if $E/F$ is separable.
For the converse, note that if $K(x_1,\ldots,x_n)$ is a finitely generated extension, and $f\in K[X_1,\ldots,X_n]$, let $\partial f/\partial x_i$ be $\partial f/\partial X_i$ evaluated at $(x_1,\ldots,x_n)$. If $f(X)\in K[X_1,\ldots,X_n]$ is a polynomial that vanishes on $(x_1,\ldots,x_n)$, then any extension $D^*$ of $D$ must satisfy
$$0 = D^*f(x) = f^D(x) + \sum\frac{\partial f}{\partial x_i}D^*x_i;$$
and this condition is also sufficient for an extension to exist. If $E/F$ is not separable, then use this necessary condition to construct a derivation that cannot be extended. 
