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27th September 2018, 21:58 #1
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General Identity of Derivative
Hello,
I know this is really stupid BUT my mind is still questioning it. I am taking the dot product of magnetic field, B with its derivative:
B.∂B/∂t ................(i)
= ∂B^{2}/∂t ........(ii)
= 2 B.∂B/∂t ..........(iii)
BUT eqn(i) ≠ eqn(iii)!!!
What is going on? Am I not thinking correctly?
Thanks in advance

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27th September 2018, 22:55 #2
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Re: General Identity of Derivative
No, you are not thinking correctly. All you wrote makes no sense.
Expression (i) is the partial derivative of variable B with respect to t, multiplied by variable B. Whatever you could ever get as the result would be variable. This is the first misconception. A derivative is just a slope of a function at a point, and nothing but a slope. Realize that this expression is the LHS that basically holds information as to what happened to the RHS  the actual expression is the RHS.
Expression (ii) is not equal to (i) because B cannot be both a constant and a variable at the same time and so the two B's cannot be multiplied together. Thus, this expression is invalid.
Expression (iii) does not follow (i) because (ii) does not follow (i) and moreso because (ii) is invalid.
The fact still remains that you can only get answers/help based on questions you ask anjd how you ask them.
Akanimo.

27th September 2018, 23:10 #3
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Re: General Identity of Derivative
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!Last edited by dzafar; 27th September 2018 at 23:16.

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27th September 2018, 23:46 #4
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Re: General Identity of Derivative
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.Last edited by Akanimo; 28th September 2018 at 00:09.

Akanimo.

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28th September 2018, 00:14 #5
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Re: General Identity of Derivative
The main mistake is thinking that x.dx is d(x^2);
That is simply because dx is not d.x; it is a new infinitesimal variable.
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30th September 2018, 22:55 #6
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Re: General Identity of Derivative
Okay.. so I figured it out.
My mistake was that I was taking B directly inside the derivative and making it a B^2. The correct way is to apply the product rule for derivative. For simplicity I am taking two different fields, B1 and B2:
∂(B1.B2)/∂t = B1.∂B2/∂t + B2.∂B1/∂t
Since, B1 = B2 = B => ∂(B1.B2)/∂t = ∂(B.B)/∂t = B.∂B/∂t + B.∂B/∂t
∂(B.B)/∂t = 2 B.∂B/∂t
Lastly, for the L.H.S, we have a vector dotted with the same vector = magnitude of the vector squared!
i.e. ∂(B.B)/∂t = ∂B^2/∂t
Hence, ∂(B.B)/∂t = ∂B^2/∂t = 2 B.∂B/∂t
I just proved none of equations (i) through (iii) are invalid!!! Let me know if you find any flaw in the explanation!
Thanks

1st October 2018, 03:38 #7
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Re: General Identity of Derivative
Good job.
I noticed that you have made a lot of assumptions and manipulations to arrive at this satisfaction. I have a few observations for you and they are majorly based on the assumptions that you have made.
1)It's not clear why you chose to represent distinct variables B1 and B2 with the same notation (placeholder) B just because they are equal at some point in time. Be wary that you are taking partial derivatives of these fields because they are variables and would change with time otherwise, you wouldn't need to since the derivative of a constant is zero. Your assumption that B1 = B2 = B concludes that they will always be equal such that for every change in B1, there will be exact same change in B2. This is a strong constraint that any 'small' difference between the two fields will violate. Assumptions such as B1 = B2 = B are usually made in situations where the placeholders (B1, B2, and B in this case) are constants rather than variables.
2)Taking a closer look, you will see that the expression B.ӘB/Әt in Post #1 is not the same as B.ӘB/Әt + B.ӘB/Әt in Post #6 but you have assumed them to be equal. In fact B.ӘB/Әt = (1/2)*(B.ӘB/Әt + B.ӘB/Әt) so expression (ii) in Post #1 does not follow expression (i).
3)You assumed that B1.ӘB2/Әt + B2.ӘB1/Әt = 2*B2.ӘB1/Әt = 2*B1.ӘB2/Әt when in reality it is not. The assumption cannot be conclusive just because B1 = B2. They still remain two distinct quantities.
Akanimo.

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1st October 2018, 05:18 #8
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Re: General Identity of Derivative
Let me know if you find any flaw in the explanation!
The oprerator d/dt working on a constant gives zero. The rest will follow automagically.

31st October 2018, 22:42 #9
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Re: General Identity of Derivative
d(f^2(x)/dx = 2f(df/dx) >
B(dB/dx) = 1/2(d(B^2)/dx)
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