# General Identity of Derivative

1. ## 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)
= ∂B2/∂t ........(ii)
= 2 B.∂B/∂t ..........(iii)

BUT eqn(i) ≠ eqn(iii)!!!

What is going on? Am I not thinking correctly?

Thanks in advance •

2. ## Re: General Identity of Derivative Originally Posted by dzafar 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)
= ∂B2/∂t ........(ii)
= 2 B.∂B/∂t ..........(iii)

BUT eqn(i) ≠ eqn(iii)!!!

What is going on? Am I not thinking correctly?

Thanks in advance
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. 3. ## 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! •

4. ## Re: General Identity of Derivative Originally Posted by dzafar 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. •

5. ## 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.

1 members found this post helpful. 6. ## 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 7. ## 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 BB/Әt in Post #1 is not the same as BB/Әt + BB/Әt in Post #6 but you have assumed them to be equal. In fact BB/Әt = (1/2)*(BB/Әt + BB/Әt) so expression (ii) in Post #1 does not follow expression (i).

3)--You assumed that B1B2/Әt + B2B1/Әt = 2*B2B1/Әt = 2*B1B2/Әt when in reality it is not. The assumption cannot be conclusive just because B1 = B2. They still remain two distinct quantities. •

8. ## Re: General Identity of Derivative

Let me know if you find any flaw in the explanation!
Just remember that d/dt is an operator; the operand must be a function. The result is also a function.

The oprerator d/dt working on a constant gives zero. The rest will follow automagically. 9. ## Re: General Identity of Derivative

d(f^2(x)/dx = 2f(df/dx) -->

B(dB/dx) = 1/2(d(B^2)/dx) --[[ ]]--