# Derivative of a constant scalar field at a point

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## Summary:

The Derivative of a scalar field at a point is normally described as the "cotangent vector of the field at that point". Can we define the derivative of the field at a point, for a constant scalar field, as the "linear density" of the field at the point?
Wikipedia defines the derivative of a scalar field, at a point, as the cotangent vector of the field at that point.

In particular;

The gradient is closely related to the derivative, but it is not itself a derivative: the value of the gradient at a point is a tangent vector – a vector at each point; while the value of the derivative at a point is a cotangent vector .

Now if a scalar fields value varies with x according to F = f(x) then the derivative is clearly f` (x) which may be different from zero (depending in the form of f(x)).

If the field has a constant value, over x, e.g Field = 4x. Can we say that the derivative = 4 and represents the "linear density" of the Field along the x axis ? In such a case the Gradient will be a vector (having magnitude = 4) being coincident with the x-axis. The Directional derivative of the field, at that point, will only exist along the x-axis (except as we may define values for the Field along the Y and Z axis) and will also have, in this case, a constant value of 4 in the direction of the x-axis.

I am only accustomed to thinking of the above in the case where f(x) is a function whose derivative is not a constant value.

Hope this question makes sense ? If I am correct then the answer to the above is trivial but will clarify things in my own mind.

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fresh_42
Mentor
Consider ##x \longmapsto x^3##.

Then the ##3x_0^2## represents the slope of the tangent at ##x_0##.

But at the same time it is the linear function ##\nabla_{x_0}## in
\begin{align*}f(x_0+v)&=(x_0+v)^3\\&=x_0^3+3x_0^2v+3x_0v^2+v^3\\
&=(x_0)^3 + 3x_0^2\cdot v+v^2(3x_0+v)\\&=f(x_0)+ \nabla_{x_0} \cdot v + r(v^2)
\end{align*}
which is "multiply the direction by ##3x_0^2##": ##v\longmapsto 3x_0\cdot v##.

So one perspective is the evaluated value at the point to construct the tangent, and the other is the linear function ## \nabla\, : \, x_0 \longmapsto L(3x_0^2)## where ## \nabla## maps a point to a linear function, namely the multiplication ## L(3x_0^2)\, : \,v \longmapsto 3x_0^2\cdot v##.

One is a vector: ##3x_0^2\cdot \vec{v}## and one a covector, a function: ##x\longmapsto L(3x^2)##.

Maybe it's better explained here:
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

Hi Fresh-42