Given a gauge transformation $u$ for a vector bundle $E\to M$, and a connection $\nabla$, we define: $$u^*\nabla = u\nabla(u^{-1}s)$$
I want to prove the well-known identity $$u^*\nabla = \nabla -(d_{\nabla}u)u^{-1}$$
A passage in the proof I'm following starts by considering: $$ \nabla s=\nabla(uu^{-1}s)=(d_{\nabla}u)(u^{-1}s)+u\nabla(u^{-1}s)$$
why are we applying the Leibniz rule to the composition $u^{-1}us$? I mean Why are we considering it as a product?