Directly get tangent vector of Bloch sphere from quantum state (qubit)?

We know that Bloch sphere is a good way to represent a qubit(two energy quantum systems). Now I want to know the tangent vector in Bloch sphere, e.g. for states $\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1\\ e^{i\varphi}\\ \end{array} \right)$ , or equivalently with $x,y,z$ coordinate: $\left( \begin{array}{c} \cos\varphi\\ \sin\varphi\\ 0\\ \end{array} \right)$ . We can calculate the tangent vector by $\partial _{\varphi}\left( \begin{array}{c} \cos\varphi\\ \sin\varphi\\ 0\\ \end{array} \right) =\left( \begin{array}{c} -\sin\varphi\\ \cos\varphi\\ 0\\ \end{array} \right)$ . My question is, is there a way to calculate a quantity similar to $\left( \begin{array}{c} -\sin\varphi\\ \cos\varphi\\ 0\\ \end{array} \right)$ without refer to $x,y,z$ coordinates? Because I want to see what the tangent vector correspond to $n$ -qubits instead of single qubit case, in that case, we can't seek help from $x,y,z$ coordinates.