Simple question on the derivative of base frame

In summary, the rotating frame has a time derivative that is different from the inertial reference frame. The only way to figure out how to apply the product rule to a rotating frame is to use the coordinate functions that started out as derivatives of coordinate functions in the inertial reference frame.
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TL;DR Summary
The derivative of a vector in a rotating frame (or even inertial)
I apologize: despite my verbosity, this is, I hope, a simple question.)

Consider the following relationship between a rotating reference frame and an inertial reference frame (both Bold), through a rotation matrix:
(the superscript is to designate the rotating frame e(1) and the I is for the inertial)

e(1)=eIR(1)

Now suppose I wish the time derivative of the rotating frame.
(I do not know how to put the dot overhead, sorry about this)
(The dot shoud be above the first e and above the R)

.e(1)=eI.R(1)

Now, did you see how I did not take the time derivative of the base frame?

Yes, PHYSICALLY, I know it is not rotating, so I do not have to do that.

But suppose I am a stubborn person (I am) who has memorized the "product rule" in calculus -- the first times the derivative of the second + the second times the derivative of the first.

And, now I see the product of a Rotation matrix (functions -- sine, cosine) and this bold base frame: eI

If I apply the rule, strictly, I put the dot over it and assert the result is zero.

But suppose I want to be stubborn and say: "but that base frame is not a function! I only know how to use the product rule for functions"

How do I answer myself?

The ONLY guess I can make is the following:

We began with coordinate functions: x1(1), x2(t) and x3(t)

We take time derivatives as we move a point, P(x1,x2, x3)

∂P(x1,x2, x3)/∂x1,
∂P(x1,x2, x3)/∂x2,
∂P(x1,x2, x3)/∂x3

Then, we decide to create the frame by dropping the point P, notation and looking at only the partial notation

e(1)≡∂/∂x1,
e(2)≡∂/∂x2,
e(3)≡∂/∂x3

We do this like Ted Frankel did on page (3) of this:
http://www.math.ucsd.edu/~tfrankel/the_geometry_of_physics.pdf

And while we have a frame, its three axes "began their lives as derivatives of coordinate FUNCTIONS, so we CAN use the product rule."

Can anyone advise a better way?

For example, suppose BOTH frames are moving and we want the derivative. We have to use the product rule and get the following sequence.

e(2)=e(1)R(2/1)

(and please forgive me for the placement of the dot--all dots should be directly overhead)

.e(2)= .e(1)R(2/1) + e(1).R(2/1)

Again, how do I know (in my ignorance and stupidity), to apply a product rule (proven for functions) to a base frame? (I have no issue, by the way, with the matrix structure formulation of the time derivative -- that is not an issue).

(And if you can tell me how to get the dot directly overhead, that would be nice, too.)
 
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Related to Simple question on the derivative of base frame

What is a base frame?

A base frame is a reference frame used in mathematics and physics to describe the position and orientation of an object. It is typically represented by a set of three orthogonal axes (x, y, z) and is used as a starting point for measuring the movement or changes in an object's position.

What is the derivative of a base frame?

The derivative of a base frame is a mathematical concept that describes the rate of change of the base frame with respect to another frame or variable. It is commonly used in calculus and physics to calculate the instantaneous rate of change of a function or object's position.

How do you calculate the derivative of a base frame?

The derivative of a base frame can be calculated using the chain rule, which involves taking the derivative of each component of the base frame with respect to the variable or frame of reference. This can be done using basic calculus techniques such as differentiation and integration.

What is the purpose of calculating the derivative of a base frame?

The derivative of a base frame is used to describe the instantaneous rate of change of an object's position or orientation. It is important in physics and engineering for understanding the motion and behavior of objects, and in mathematics for solving equations and optimizing functions.

Can the derivative of a base frame change over time?

Yes, the derivative of a base frame can change over time if the object's position or orientation is changing. This change can be represented by a function or equation that describes the relationship between the base frame and the variable or frame of reference. Differentiating this function will give the derivative of the base frame at a specific point in time.

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