why z-transform differentiation is needed?
You just wrote down the $\mathcal{Z}$-transform of $\left(\frac12\right)^nu[n]$, but you need the $\mathcal{Z}$-transform of $n\left(\frac12\right)^nu[n]$ (note the factor $n$).

In order to find that $\mathcal{Z}$-transform you can use the differentiation property (see this table):

$$\mathcal{Z}\big\{nx[n]\big\}=-z\frac{dX(z)}{dz}\tag{1}$$