polynomials - Prove two bases are dual in a finite field.

Let K be a finite field, $F=K(\alpha)$ a finite simple extension of degree $n$, and $ f \in K[x]$ the minimal polynomial of $\alpha$ over $K$. Let $\frac{f\left( x \right)}{x-\alpha }={{\beta }_{0}}+{{\beta }_{1}}x+\cdots +{{\beta }_{n-1}}{{x}^{n-1}}\in F[x]$ and $\gamma={f}'\left( \alpha \right)$.

Prove that the dual basis of $\left\{ 1,\alpha ,\cdots ,{{\alpha }^{n-1}} \right\}$ is $\left\{ {{\beta }_{0}}{{\gamma }^{-1}},{{\beta }_{1}}{{\gamma }^{-1}},\cdots ,{{\beta }_{n-1}}{{\gamma }^{-1}} \right\}$.

I met this exercise in "Finite Fields" Lidl & Niederreiter Exercises 2.40, and I do not how to calculate by Definition 2.30. It is

Definition 2.30 Let $K$ be a finite field and $F$ a finite extension of $K$. Then two bases $\left\{ {{\alpha }_{1}},{{\alpha }_{2}},\cdots ,{{\alpha }_{m}} \right\}$ and $\left\{ {{\beta }_{1}},{{\beta }_{2}},\cdots ,{{\beta }_{m}} \right\}$ of $ F$ over $K$ are said to be dual bases if for $1\le i,j\le m$ we have $T{{r}_{{F}/{K}\;}}\left( {{\alpha }_{i}}{{\beta }_{j}} \right)=\left\{ \begin{align} & 0\;\;\text{for}\;\;i\neq j, \\ & 1\;\;\text{for}\;\;i=j. \\ \end{align} \right.$

I think $\gamma =\underset{x\to \alpha }{\mathop{\lim }}\,\frac{f(x)-f{{(\alpha )}_{=0}}}{x-\alpha }={{\beta }_{0}}+{{\beta }_{1}}\alpha +\cdots {{\beta }_{n-1}}{{\alpha }^{n-1}}$.

How can I continue? The lecturer did not teach the "dual bases" section.