Let $v\in \mathbb{H}$ and $q:t\in\mathbb{R}\rightarrow q(t)\in \mathbb{H}$. That $q(t) \neq 0$ for all $t\in \mathbb{R}$ and $q^{-1}(t) = \frac{1}{q(t)}$. So don't confuse $q^{-1}(t)$ with inverse function(as pointed by rschwieb).
Than we have limit:
$$\lim_{\Delta t \rightarrow 0} \frac{1-q(t+\Delta t) q^{-1}(t)}{\Delta t} =\lim_{\Delta t \rightarrow 0} \frac{q(t)-q(t+\Delta t)}{\Delta t} q^{-1}(t) = - q'(t)q^{-1}(t) $$
But I have trouble to find limit of this:
$$\lim_{\Delta t \rightarrow 0} \frac{v-q(t+\Delta t) q^{-1}(t)vq(t) q^{-1}(t +\Delta t)}{\Delta t} = ?$$
Does it even exist? Can it be writen i terms of $v,q,q'$?
Context: I developed some system of recurrence equations. See And I want to find corresponding differential equations:
This is the system: $i \in \{1,\dots,n\}$ \begin{align*} v_i(t_{n+1})& = p_i(t_{n+1})-p_0(t_{n+1}) = R_0(t_n,\Delta t) v_i(t_n) R_0^{-1}(t,\Delta t)\\ p_0(t_{n+1}) &= p_0(t_n) + \sum_{i=1}^n v_i(t_n) - R_i(t_n,\Delta t) v_i(t_n) R_i^*(t_n,\Delta t) \\ q_0(t_{n+1}) &= \{R_i(t_n,\Delta t)\}_i q_0(t_n) \\ p_i(t_n) &= p_0(t_n) + R_0(t_0,t_n-t_0)v_i(t_0)R_0^{-1}(t_0,t_n-t_0) \\ R_i(t,\Delta t) &= q_i(t+\Delta t) q^{-1}_i(t)\\ \{a_i\}_i &=\frac{\sum_{ \sigma \in \Pi_n } a_{\sigma(1)} \dots a_{\sigma(n)}}{||\sum_{ \sigma \in \Pi_n } a_{\sigma(1)} \dots a_{\sigma(n)}||} \end{align*}
here is how I came up with it
This can be evaluated with an invocation of the product rule for quaternion functions:
$$(fg)'(t) = f'(t)g(t)+f(t)g'(t)$$
namely by considering $f(\tau) = -q(\tau)q^{-1}(t)$ and $g(\tau) = vq(t)q^{-1}(\tau)$, deriving at $\tau = t$. This means the result is $-q'(t)q^{-1}(t)v-vq(t)(q^{-1})'(t)$. Note that we have assumed $q^{-1}$ to be differentiable.
A derivation of the product rule goes along the standard lines, exploiting the identity:
$$f(t+\Delta t)g(t+\Delta t) - f(t)g(t) = f(t+\Delta t)\left(g(t+\Delta t)-g(t)\right)+\left(f(t+\Delta t)-f(t)\right)g(t)$$