I agree that it's hard to understand. So ( S _D F ) is the derivative of F restricted to S ?
By the way, it's strange that in both versions, the product in the second summand is commuted. I mean: one has (fg)' = f'g + fg' (since \C is commutative, you can certainly write it as (fg)' = f'g + g'f, but it's weird). Benoît -- You received this message because you are subscribed to the Google Groups "Metamath" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/metamath/fbd68af8-4edf-4260-b917-d069caa514d6%40googlegroups.com.
