Does a derivative divided by its modulus value give an unit vector?

Question

Consider the curve $y = 2x^2$.

The derivative at any point on the curve is $$ \frac{dy}{dx} = 4x. $$

since, $f\ '\ ( x)$ at a point $(x)$ gives the tangent vector, we get $\overrightarrow{t} = \frac{dy}{dx}$

therefore the unit vector, $\hat{t} = \frac{\overrightarrow{t}}{\| \overrightarrow{t} \|}$ = $\frac{\frac{dy}{dx}}{\|\frac{dy}{dx}\|} = \frac{4x}{\|4x\|}$

therefore, for $x = 4$, $\hat{t} = 1$

while for $x = -2$, $\hat{t} = -1$

and, for $x = 0$, $\hat{t} = 0$

QUESTION 1:

is my assumption that a derivative divided by its modulus value give a unit vector correct?

QUESTION 2:

If yes, then in which direction is the unit vector pointing? if false, then, what does $\frac{dy}{\|\frac{dy}{dx}\|}$ represent?

Answer 1

Thinking of f(x) as a vector in the vector field is okay and the direction can be chosen as per your convenience (since it's one dimensional, or, it assigns just one value to every point in xy plane, that being -1/0/1). Since it tells the sign of derivative and thus, the tendency of change in the function's output, it could be that it points in the positive y direction for +1 and opposite for - 1, it being a zero vector for 0, since it's indicating of the corresponding tendency of the function (here the dimension being the y axis).

Update : as the vector z is defined, it's only assigned a magnitude. Thus it could be any vector with magnitude associated with that point in the xy plane. Therefore, thinking of the unit vector z does not make much sense, or, it's not defined well. You could assign a particular direction to it in the definition or think of gradient vector but it would mean the unit vector signifies the direction where there's maximum change in function (towards x axis)

Answer 2

Strictly speaking, $f'(x)$ is the derivative of $f(x)$, so it is just another function.

If you get into higher mathematics, the word "vector" takes on a very general meaning that encompasses many things you will not see in high school or first-year calculus. So I will not argue whether $f'(x)$ can be a "vector" or not. I will simply take note that you are talking about a "tangent vector" in relation to a function from $\mathbb R$ to $\mathbb R$, which almost certainly means you are looking at graphs of functions in a Cartesian plane with two dimensions, and you want a vector in the two-dimensional vector space over that plane.

If you have a graph in a Cartesian plane on which a curve is defined by $y = f(x),$ then at any particular point $(x_1,f(x_1))$ on the curve you can construct a tangent line. If the tangent line is parallel to the $y$ axis there is no derivative at that point, but otherwise the derivative at that point, $f'(x_1),$ is the slope of the tangent line.

To obtain a tangent vector you just need to take the vector between any two points on the tangent line. In fact, since vectors in this context have only magnitude and direction, not any particular start or end points, you can take the vector between any two points on any line parallel to the tangent line.

Such a vector will have both an $x$ coordinate and a $y$ coordinate. You can then use the formula $\frac v{\lVert v\rVert}$ to obtain a unit vector, again with two coordinates.