Is $\frac{\rm d}{\rm d\omega(t)}\int_{t_0}^t\omega(t')\rm dt'=\int_{t_0}^t\frac{\rm d \omega(t')}{\rm d\omega(t')}dt'=t_0-t?$

Question

I'm trying to calculate an error propagation, but the expression in the most LHS of the equation in the title crops up. Are you allowed to simply exchange the order so that the operations cancel?

$\omega(t)$ is a simple angular velocity that is a dataset obtained from an experiment with a pendulum.

EDIT: Apparently it cannot be done. I only know $\omega(t)$ numerically (it's a data-array) - any ideas for how to calculate an error propagation that involves this term?

I found this example that might be helpful, but I'm a bit in the deep end here, and Leibniz's rule doesn't seem to apply here.

If this error propagation could somehow be obtained numerically, that would also be very helpful!

Any suggestions?

Thanks!

Answer 1

I would say that

$$\frac {\textrm d} {\textrm d \omega (t)} \int \limits _{t_0} ^t \omega(t') \ \textrm d t' = \omega (t) \frac {\textrm d t} {\textrm d \omega (t)} = \frac {\omega (t)} {\omega ' (t)} .$$

Now, having $t \mapsto \omega (t)$ numerically, compute $t \mapsto \omega' (t) = \frac {\omega (t+h) - \omega(t)} h$ numerically (for some chosen, small, fixed $h$) and this should be it.

The results used above are:

$$\frac {\textrm d} {\textrm d t} \int \limits _{f(t)} ^{g(t)} \varphi (t') \ \textrm d t' = \varphi \Big( g(t) \Big) g'(t) - \varphi \Big( (f(t) \Big) f'(t)$$

and, obviously,

$$\frac {\textrm d \omega} {\textrm d t} = \frac 1 {\frac {\textrm d t} {\textrm d \omega}}$$

(under the usual assumptions involving the existence of fractions etc.)