Can someone help me solve this problem or tell me what I need to proof?
Suppose $V = P_4(\mathbb{R})$ and let $B = \{1, x, x^2, x^3, x^4\}$ be the standard basis. Prove that the map $f(p(x)) = p′(3)$ is in $V^*$. Express it as a linear combination of the dual basis $\mathcal{B}$.
(Hint: Do note that $B^*$ is not $B$, see 1.c.i).
Let $B=\{b_i=x^i : 0\leq i\leq 4\}$ be the standard basis for $V=P_4(\mathbb{R})$.
Then the dual basis $B^*=\{b^i : 0\leq i\leq 4\}$ of $V^*$ has elements as follows: $$ b^i\colon V\to\mathbb{R} \text{ defined by } b^i(b_j)=\begin{cases} 1 & \text{if $i=j$} \\ 0 & \text{if $i\neq j$} \end{cases} $$
It is easy to show that $f\in V^*$ so I leave it for you. (Hint: differentiation is linear.)
Now the linear map $f\colon V\to\mathbb{R}$ is defined by $f(p(x))=p'(3)$, in other words, $$ f(a_0b_0+a_1b_1+a_2b_2+a_3b_3+a_4b_4) = a_1+(2a_2)\cdot3+(3a_3)\cdot3^2+(4a_4)\cdot3^3 $$ where $a_i\in\mathbb{R}$ for all $0\leq i\leq 4$.
Therefore $f=b^1+(2\cdot3)b^2+(3\cdot3^2)b^3+(4\cdot3^3)b^4$ in the dual space $V^*$.