Okay!! I know what I wrote DOES make sense. B is NOT a variable. It is a VECTOR FIELD!!!
Once you understand that it is a vector, then you would see that vector dot the same vector is equal to the MAGNITUDE squared of the vector!! and hence (ii) is correct!
You've gone out of context. It doesn't matter what it is. Whatever it is falls into either
variable or
constant. Here, it is a variable, otherwise you wouldn't be considering the derivative of a constant since you would always get
0 as the result.
The main point here is that you cannot multiply y.(dy/dx) to get d(y^2)/dx. In otherwords, y.dy/dx=y.dy/dx.
I said something about expression (ii) being invalid. This is because I considered it to be the LHS of an expression with the RHS being an expression with variables
B,
t,<some_constant>, and/or <some_other_variable(s)>. Now, assuming B^2 is not the LHS but the RHS such that you have an expression that goes like this:
y = B^2 and you are finding its derivative with respect to
t, that derivative would be zero. You wouldn't be computing a derivative that you already know would result to
0 -- so expression (ii) is still invalid. However, B^2 is not the RHS, else you would not have been able to have it written as B.dB/dx
What are you trying to do? We can help you out.