Suppose that normal derivative = [itex]\nabla[/itex]g . n = dg/dn,(adsbygoogle = window.adsbygoogle || []).push({});

then f [itex]\nabla[/itex]g . n = f dg/dn

[I used . for dot product]

But how is this possible?

For f [itex]\nabla[/itex]g . n, I would interpret it as (f [itex]\nabla[/itex]g) . n

But f dg/dn = f ([itex]\nabla[/itex]g . n) which is DIFFERENT (note the location of brackets)

I recall that ifu,vare vectors, a is a scalar constant, then (au) .v=a(u.v) =u. (av), but in our case, f is a FUNCTION!!! So this rule cannot be applied.

f [itex]\nabla[/itex]g . n = f dg/dn seems to be suggesting that f is a scalar constant, but consider the following case:

Let div F=[itex]\nabla[/itex] . F

If f were really a scalar constant, then by the rule, div (f [itex]\nabla[/itex]g) = [itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . ([itex]\nabla[/itex]g)) which is WRONG becuase we know that in general div(fG)=f divG+ ([itex]\nabla[/itex]f) .Gwhere f is function and G is vector field. But if this is wrong, then HOW can you justify that (f [itex]\nabla[/itex]g) . n = f ([itex]\nabla[/itex]g . n)?

I seriously can't understand this...there are two seemingly contradicting ideas crashing in my mind...

Would someone be nice enough to clear my doubts? Thanks!

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# Functions & vector fields

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