TITLE: How does the chain rule work for functions from vectors to vectors?
QUESTION [5 upvotes]: Suppose I have a function:
$$ \vec{s} = \vec{f}\left(\vec{\theta}\right)$$
and a derivative:
$$ \vec{v} = \frac{\mathrm{d} \vec{s}}{\mathrm{d} t}$$
How do I apply the chain rule?
For simplicity lets call $\omega = \frac{\mathrm{d} \vec{\theta}}{\mathrm{d} t}$
I think the chain rule should be something along the line of:
$$ \vec{v} = \vec{\omega} \times \nabla_{\theta} \vec{f}\left(\theta\right) $$
but I don't know the exact rule.
I think I may have to use matrices and more complicated derivatives like the Jacobian.

REPLY [3 votes]: Consider a particular component $v_k$ of the velocity vector $\vec{v}(\vec{\theta})$. Then the chain rule yields 
$$\displaystyle v_k = \frac{ds_k}{dt}= \sum_{j=1}^3 \frac{\partial s_k}{\partial \theta_j}\frac{d\theta_j}{dt} =\sum_{j=1}^3 \frac{\partial s_k}{\partial \theta_j}\omega_j.$$ If we understand $\nabla_{\vec{\theta}} F=\sum_{j=1}^3 \frac{\partial F}{\partial \theta_j}\hat{e}_k,$ the above may be written as $v_k=\nabla_{\vec{\theta}}s_k\cdot \vec{\omega}$ i.e. $\vec{v}=\nabla_{\vec{\theta}}\vec{s}\cdot \vec{\omega}.$ This reflects the fact that the first term is a rank 2 tensor, and so the use of the dot product (instead of the cross product) is warranted in order to obtain a vector.