Let $V = \{\ p\ \in\ \mathbb{F}_2[x]\ \ \ |\ \ \deg(p)\ \leq\ 4\}$ be the vector space of polynomials with degree no greater than 4 and with coefficients in the finite field $\mathbb{F}_2$. Consider the linear map given by differentiation: $$D: \ V \to V$$ $$p(x) \to \frac{dp}{dx}$$ Write down the matrix of $D$ with respect to the standard monomial basis $1,\ x,\ x^2,\ x^3,\ x^4 $ of $V$
Without thinking about the field $\mathbb{F}_2$ I would do:
$D(1)=0=0\cdot 1+0\cdot (x)+0\cdot (x^2)+0\cdot (x^3)+0\cdot (x^4)$ $D(x)=1=1\cdot 1+0\cdot (x)+0\cdot (x^2)+0\cdot (x^3)+0\cdot (x^4)$ $D(x^2)=2x=0\cdot 1+2\cdot (x)+0\cdot (x^2)+0\cdot (x^3)+0\cdot (x^4)$ $D(x^3)=3x^2=0\cdot 1+0\cdot (x)+3\cdot (x^2)+0\cdot (x^3)+0\cdot (x^4)$ $D(x^4)=4x^3=0\cdot 1+0\cdot (x)+0\cdot (x^2)+4\cdot (x^3)+0\cdot (x^4)$
resulting in the matrix: $$\begin{bmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 3 & 0\\0 & 0 & 0 & 0 & 4\\0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$$ My question is, how does this change with respect to the field $\mathbb{F}_2$? I am not too confident on how fields work so any extra explanation would be appreciated.
In $\Bbb F^2$, the elements $2$, $3$ and $4$ just "don't exist". In other words, your work is correct once you use that $3=1$ and $4=2$. For completeness: we have $$\begin{align} D(1) &= 0 \\ D(x) &= 1 \\ D(x^2) &= 0 \\ D(x^3) &= x^2 \\ D(x^4) &= 0,\end{align}$$so $$D = \begin{pmatrix} 0 & 1 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{pmatrix}.$$