This paper discusses an important property of the derivative of a unit vector, which connects the geometric meaning of differentiation of vector functions with algebraic transformations. Particular attention is paid to the global vector \(\R\), defined through the vector transformation of a scalar function, and its role in unit space. It is shown that differentiation of this vector preserves orthogonality between the vector and its derivative, which finds direct application in the analysis of curves and fields, as well as in more complex mathematical structures, including hyperbolic numbers and generalized spaces.

The work is based on the previously introduced concepts of vector transformation of scalar functions and conjugate functions in the extension of hyperbolic numbers, which allows us to unify approaches and perform calculations in a single notation. The theoretical calculations are accompanied by verification on a specific example illustrating the correctness of the key formula through the expansion of sine into a power series.

Thus, the goal of the work is to rigorously derive and justify the property of the derivative of a unit vector, formalized through the scalar product of a vector function and its derivative. The result is of interest for further study of vector spaces, methods of differential geometry, field theory and related disciplines.

In the theory of a unit space, the vector \(\R\) is called global. We will use this term here as well. It has unique properties and is defined as follows: \[\tag{1} \R = \R(x) = {\f(x) \over f(x)} \] Here: \(\f(x)\) - vector transformation of the scalar function \(f(x)\).

From the main property of vector transformation of functions, we know that \[\tag{2} \f(x) \cdot \f(x) = f^2(x) \] Moreover, in the general case, when a minus sign appears under the root during the transformation, we must multiply the conjugate vector by the initial one according to the rules of hyperbolic numbers \[\tag{3} \fo(x) \cdot \f(x) = f^2(x), \] approximately as it is done in quantum mechanics with a complex modulus. We will not point this out every time, but we will imply it.

From this follows another important property of the unit space: \[\tag{4} \R \cdot \R = 1 \] If we now take the derivative with respect to \(x\), we get zero as a result: \[\tag{5} (\R \cdot \R)_x^{'} = (1)_x^{'} = 0 \] And this means that the derivative vector is always perpendicular to the original vector: \[\tag{6} \R \cdot \R^\!{'} = 0 \] This is also one of the properties of the unit space.

To simplify perception, we will assume that all derivatives are related to \(x\). For this purpose, we will simplify the notation of functions: \[\tag{7} \f = \f(x), \quad f = f(x) \] Now we can take the derivative of the global vector \[\tag{8} \R^\!{'} = {\fp f - \f\, f^\!{'} \over f^2} \] and multiply it from the left by the original vector. From (6) we know that this will give zero \[\tag{9} \R \cdot \R^\!{'} = {\f\, \fp f - \f\, \f\, f^\!{'} \over f^3} = 0 \] Leaving only the numerator \[\tag{10} \f\,\fp f = \f\, \f\, f^\!{'} \] and substituting property (2) there \[\tag{11} \f\, \f = f^2 \] we get the final formula: \[\tag{12} \f\, \fp = f f^\!{'} \]

Let's check formula (13) on a specific example. For this, we will take a vector transformation for the sine to also show the actions with the hyperbolic unit. Let's use the ready-made solution, but write it in the following form: \[\tag{14} \Sin(\v) = \sumn1 \j_{2n}\, (\i)^{n-1}\, {\a^{n} \over \sqrt{2\, (2n)!} } \\ \v = {\a \over 2} \] Let's take the derivative with respect to the angle \(\v\), but since this angle is equal to twice \(\a\), the result must be multiplied by 2: \[\tag{15} \Sin(\v)_{\v}^{'} = \sumn1 \j_{2n}\, (\i)^{n-1}\, {2n\, \a^{n-1} \over \sqrt{2\, (2n)!} } \] Now the scalar newe multiply the vector sine and its derivative according to formula (13): \[\tag{16} \overline{\Sin}(\v) \cdot \Sin(\v)_{\v}^{'} = \sumn1 (\oi)^{n-1} (\i)^{n-1}\, {2n\, \a^{2n-1} \over 2\, (2n)! } = P \] According to the rules of operations with the hyperbolic unit we know that \[ (\oi)^{n-1} (\i)^{n-1} = (-1)^{n-1} \] Then expression (16) can be written as follows: \[\tag{17} P = \frac12 \sumn1 (-1)^{n-1} {\a^{2n-1} \over (2n-1)! } = \frac12 \sumn0 (-1)^{n} {\a^{2n+1} \over (2n+1)! } \] It remains to do the same with the product of the scalar sine and its derivative, the cosine: \[\tag{18} \sin(\v) \cos(\v) = \frac12 \sin(\a) = P \] It is obvious that the values ​​of formulas (17) and (18) are equal if we look at the Maclaurin series for sine [1]. This means: \[\tag{19} \<\kern1pt \Sin(\v), \Sin(\v)_{\v}^{'} \> = \sin(\v) \cdot \sin(\v)_{\v}^{'} \] The check of the key formula (13) was successful.

The paper shows that the derivative of a unit vector is always perpendicular to the vector itself, which reflects the fundamental properties of normalized directions in space. A general relationship between a scalar function and its vector transformation is derived (13), confirmed by a specific example using expansion in a power series. The obtained result is an important tool for vector analysis and can be used in geometry, physics and related mathematical disciplines.