How to obtain the Jacobian $J_{\dot{x}}(\dot{q})$ of the derivatives from the "normal" Jacobian $J_{{x}}({q})$?

Question

I have a problem that arose in a kinematics context. Suppose, the Jacobian $J_{x}(q) = \frac{\partial{x}}{\partial{q}}$ of vector $x$ w.r.t. vector $q$ is known. I am interested in the Jacobian $J_{\dot{x}}(\dot{q}) = \frac{\partial{\dot{x}}}{\partial{\dot{q}}}$ of the vector $\dot{x}=\frac{\mathrm{d}{x}}{\mathrm{d}{t}}$ w.r.t. the vector $\dot{q}=\frac{\mathrm{d}{q}}{\mathrm{d}{t}}$, with $t$ being scalar.

(In the kinematics context, $x$ is a Cartesian position, $q$ is a set of joint angles, and $t$ is time, so $\dot{x}$ is Cartesian velocity and $\dot{q}$ is the vector of joint velocities.)

Is $J_{\dot{x}}(\dot{q}) = J_{{x}}({q})$?

Answer 1

Let's do it step by step $$\begin{align} \frac{\partial\dot x}{\partial\dot q}=\frac{\partial(\frac{\partial x}{\partial q}\dot q)}{\partial\dot q}=\frac{\partial^2x}{\partial\dot q\partial q}\dot q+\frac{\partial x}{\partial q} \end{align}$$ In kinematics, $q=q(x)$ is only a function of position, or $x=x(q)$ is irrelevant to $\dot q$. We can conclude that $\frac{\partial^2x}{\partial\dot q\partial q}=0$. So your conclusion is right.