If char K=0 , then every irreducible polynomial is separable The following statement was left as exercise in my Field Theory class.

Consider a field K and f $\in K[x] $ an irreducible polynomial. Prove that f is separable in some splitting field of f over K if every root of K is simple root which is equivalent to saying that derivative of f is non-zero.

I am not able to understand how the definition of separability of a polynomial which is the fact that f is called a separable polynomial over K if all the roots of f are distinct in algebraic closure of K relates to saying that derivative of f is non -zero?
and hence I was unable to make any progress.
 A: Let $f\in K[X]$ be irreducible with a multiple root $\lambda$ in some splitting field. By the product rule of derivatives, $X-\lambda$ divides both $f$ and $f'$. In particular, $\gcd(f,f')\in K[X]$ is not constant. Since $f$ is irreducible, it follows that $\gcd(f,f')=f$. However, $f'$ has smaller degree than $f$. So this can only hold if $f'=0$. In turn, $K$ has characteristic $p>0$ and $f=g(X^p)$ for some polynomial $g\in K[X]$.
A: 
The following statement was left as exercise in my Field Theory class.

Consider a field K and f $\in K[x] $ an irreducible polynomial. Prove that f is separable in some splitting field of f over K if every root of K is simple root which is equivalent to saying that derivative of f is non-zero.

I am not able to understand how the definition of separability of a polynomial which is the fact that f is called a separable polynomial over K if all the roots of f are distinct in algebraic closure of K relates to saying that derivative of f is non -zero?
and hence I was unable to make any progress.

I think your confusion is in which two statements are claimed to be equivalent in the formulation of the exercise, so I will try to clarify that. Basically, the statement in your exercise relates the multiplicity of the roots of an irreducible polynomial with its derivative being nonzero. Specifically:
Claim. Let $f \in K[X]$ be irreducible and nonconstant. Every root of $f$ in a splitting field $L$ for $f$ over $K$ is simple if and only if $f'$ is nonzero.
Proof. Suppose that $f$ has degree $d > 0$, with roots $\alpha_1, \dotsc, \alpha_d$ in $L$.
If every root is simple, then all the $\alpha_i$'s are distinct. Writing $f = (X - \alpha_1) \dotsm (X - \alpha_k)$, we see that $f' = \sum_{i = 1}^d \prod_{j \neq i} (X - \alpha_j)$, which is nonzero since (for instance) $f'(\alpha_1) = \prod_{j = 2}^d (\alpha_1 - \alpha_j) \neq 0$.
Conversely, suppose that $f'$ is nonzero. Suppose, for the sake of contradiction, that $f$ has a repeated root in $L$. Without loss of generality, we may assume that $\alpha_1 = \alpha_2 = \alpha$. Writing $f = (X - \alpha)^2 g$, for some polynomial $g$ over $L$, we see that $f' = (X - \alpha) \cdot [2g + (X - \alpha) g']$. Hence, $(X - \alpha) \mid \gcd(f,f')$, where the $\gcd$ is computed in $L[X]$, so $f$ and $f'$ have a common factor over $L[X]$.
But, note that the Euclidean algorithm for computing $\gcd(f,f')$ only depends on the smallest field containing the coefficients of $f$ and $f'$, which is contained in $K$. So, $\gcd(f,f') \neq 1$ over $K$, too! But, this means that $f$ is not irreducible over $K$, which is a contradiction. $\blacksquare$
Do spend a moment thinking over why the first statement in the last paragraph of the proof is true. I first came across this argument in Patrick Morandi's Field and Galois Theory (GTM 167, Springer), and was immediately struck by it. You can find it there in the first paragraph of the proof of Proposition 4.5 on page 41. (For completeness, it might be worth noting that the full proof of that proposition needs a bit more work, as described here: Doubt in proof of Proposition 4.5 from Patrick Morandi's *Field and Galois Theory*. But the part relevant for us is perfectly fine.)
