First, note that a 4th degree polynomial might not exist, for a simple example, consider finding a quadratic which has points $f(0) = 0, f(1) = 1, f'( \frac{1}{2} ) = 1 $. The first 2 conditions naturally imply that $ f'( \frac{1}{2}) = 0 $, which contradicts the third.
But otherwise, we can use a similar idea to the building blocks of Lagrange Interpolation.
Use $f_1 = B(x-x_2)(x-x_3)(x-x_4)(x-A)$, where $A$ is chosen such that $ f''' (x_5) = 0$ and $B$ is chosen such that $ f_1 (x_1) = 1$. There are no issues here. Define $f_2, f_3, f_4$ similarly.
Use $ f_5 = C (x-x_1)(x-x_2)(x-x_3)(x-x_4) $, where $C$ is chosen such that $ f'''(x_5) = 1$. The only issue here, is if $ [ (x-x_1)(x-x_2)(x-x_3)(x-x_4)]''' (x_5) = 0$, in which case such a constant does not exist. In this case, we need to tag on another linear term, to make a polynomial of degree 6.