Does anyone know a general simplification: $\frac{d\vec{r}}{d||\vec{r}||}$ in terms of the unit vector $\hat{r}$. For a linear vectorfunction $\vec{r}$ the following is true of course: $\frac{d\vec{r}}{d||\vec{r}||}=\hat{r}$.
Is the following true for every $\vec{r}$ vector: $\frac{d\vec{r}}{d||\vec{r}||}=d\hat{r}$...?
Let $\,r^2=(\vec r\cdot\vec r),\,r=\|\vec r\|,\,$ and $\,\hat r={\vec r}/{r}.\;$ Then the following derivatves are true $$\eqalign{ \frac{dr}{d\vec r} &= \hat r, \qquad \frac{d\hat r}{d\vec r} &= \tfrac{1}{r}(I-\hat r\hat r^T) \\ }$$ but the remaining variable combinations $\,\Big(\frac{d\vec r}{dr},\,\frac{d\vec r}{d\hat r},\,\frac{d\hat r}{dr},\,\frac{dr}{d\hat r}\Big)\,$ are nonsense.
Sometimes you can invert a derivative to get what you want, but the above derivatives do not have inverses. They do have pseudoinverses, so depending on the application that you have in mind, you could loosely assert something like $$\eqalign{ \frac{d\vec r}{dr} &\approx \left(\frac{dr}{d\vec r}\right)^+ =\; \hat r^+ \;=\; \hat r^T \\ \frac{d\vec r}{d\hat r} &\approx \left(\frac{d\hat r}{d\vec r}\right)^+ =\; r\,(I-\hat r\hat r^T) \\ }$$ for two of the nonsensical derivatives.