Derivative of the product of a scalar function and a vector valued function

Question

According to Berkeley Physics Course, Volume 1 Mechanics,

The time derivative of a vector valued function can be derived from the formula: $$ \mathbf{r}(t) = r(t)\mathbf{\hat{r}}(t) $$ where the scalar $r(t)$ is the length of the vector and $\mathbf{\hat{r}}(t)$ is a vector of unit length in the direction of $\mathbf{r}$. The derivative of $\mathbf{r}(t)$ is defined as: $$ \begin{equation} \tag{1} \frac{d\mathbf{r}}{dt} = \frac{d}{dt}[r(t)\mathbf{\hat{r}}(t)] = \lim_{\Delta t \to 0}\frac{r(t+\Delta t)\mathbf{\hat{r}}(t+\Delta t) - r(t)\mathbf{\hat{r}}(t)}{\Delta t} \end{equation} $$ Now we use the Taylor series expansion for each of the factors $r(t+\Delta t)$ and $\mathbf{\hat{r}}(t+\Delta t) $ and we only retain the first two terms in the series expansions of $r(t+\Delta t)$ and $\mathbf{\hat{r}}(t+\Delta t)$. The general Taylor expansion series is:

$$ \begin{equation} \tag{2} f(x) = f(x_{0}) + (x-x_0)\left[ \frac{df(x)}{dx}\right]_{x=x_0} + \frac{1}{2}(x-x_{0})^{2} \left[ \frac {d^{2}f(x)}{dx^{2}} \right]_{x=x_0} + \cdots \end{equation} $$

After substituting the first two terms of the Taylor expansion series(2) into the numerator of equation $(1)$, we obtain $$ \begin{equation} \tag{3} \left[r(t)+\frac{dr}{dt} \Delta t\right] \left[\mathbf{\hat{r}}(t) + \frac {d\mathbf{\hat{r}}}{dt} \Delta t\right] - r(t)\mathbf{\hat{r}}(t) = \Delta t \left(\frac{dr}{dt}\mathbf{{\hat{r}}} + r\frac{d\mathbf{\hat{r}}}{dt} \right) + (\Delta t)^{2}\left(\frac{dr}{dt}\frac{d\mathbf{\hat{r}}}{dt}\right) \end{equation} $$

Substituting $(3)$ in $(1)$, we obtain: $$ \begin{equation} \tag{4} \frac{d\mathbf{r}}{dt} = \frac{dr}{dt}\mathbf{\hat{r}} + r\frac{d\mathbf{\hat{r}}}{dt} \end{equation} $$

Equation $(4)$ is the general formula for the derivative of the product of a scalar function and a vector valued function.

My question is that why do we prove this formula as an estimated value? Why can't we directly use the product rule for differentiation here?

Answer 1

The author's point here is that one can treat a vector function like a scalar function when it comes to differentiation. This may be obvious but the book is meant for students with a limited experience with physics, and they could find themselves asking whether or not they can differentiate vector functions. Generally speaking, in introductory physics courses all the vector functions are just constant vectors, and therefore $d\vec{v}/dt=0$, so it's not unlikely that an inexperienced student may not think about differentiating $\vec{r}$ (crucial when written in non cartesian coordinate systems).