If $R$ is a commutative ring, then the set of all polynomials with coefficients in $R$ is denoted by $R[x]$.
$\mathbb{I}_p[m]$ is the integers mod $m$.
When $p$ is a prime, we will usually denote the field $\mathbb{I}_p[x]$ by $\mathbb{F}_p[x]$.
If $R$ is a commutative ring and $f(x)=\sum^n_{i=0}a_ix^i\in R[x]$ has degree $n\geq1$, define its derivative $f'(x)\in R[x]$ by \begin{equation*} f'(x)=a_1+2a_2x+3a_3x^2+...+na_nx^{n-1}; \end{equation*} if $f(x)$ is a constant polynomial, define its derivative to be the zero polynomial.
My efforts:
$x^p=x,x^{2p}=x^2,f(x)=ax^2+bx+c$.
On one hand, $f'(x)=2pax^{2p-1}+pbx^{p-1}$.
On the other hand, $f'(x)=2ax+b$.
Since both are correct, $px^{2p-1}=x$ and $px^{p-1}=1$. So $px^{p-1}x=x$, $px^p=x$, $px=x$, $p=1$.
Where am I wrong?