I was wondering if there was a formula for the derivative of a norm (||V(t)||)' (where V(t) is some vector function from $\mathbb{R}$ to $\mathbb{R}^{n}$).
Using the dot product rule, I got $$(||V(t)||^{2})' = (V(t) \cdot V(t))' = 2V(t) \cdot V'(t)$$ and, letting g(t) = ||V(t)||, $$ (||V(t)||^{2})' = (g(t)^{2})' = 2g(t)g'(t) = 2||V(t)||*(||V(t)||)'$$
Combining both equations gives $$(||V(t)||)' = \frac{V(t) \cdot V'(t)}{||V(t)||}$$
I just wanted to check with someone if this is correct, or if I've done something wrong somewhere. Thank you!
Edit: Forgot to put in dots for some of the dot products
The reasoning is correct but you should end up with a scalar so your result is wrong.
I will denote $\phi(t)=\| \mathbf{v}(t) \|$. Thus $\phi^2=\| \mathbf{v}(t) \|^2 =\mathbf{v}(t)^T \mathbf{v}(t)$. It follows the equality $$2\phi d\phi = 2 \mathbf{v}(t)^Td\mathbf{v}(t) = 2 \mathbf{v}(t)^T\frac{\partial \mathbf{v}(t)}{\partial t} dt $$ From here, the derivative wrt time $t$ is $$ \dot{\phi}(t)=\frac{1}{\phi} \mathbf{v}(t)^T \dot{\mathbf{v}}(t) $$