Note: This is a math question by heart.
In physics the equation for work is $W = \int{\tau}\space{d}\theta$ and equation for power, $P$, is $P = \tau\omega$ ($\omega = \frac{d\theta}{dt}$). Because $P$ is also equal to the derivative of W in respect to time ($P = \frac{dW}{dt}$). What I suspect but not entirely confident is that when you take the derivative of the first function of W in respect to time, you can use the $dt$ differential to change the differential $d\theta$ in the following steps: $$1)\space W = \int\tau\space{d}\theta$$ $$2)\space\frac{d}{dt}\left[W\right] = \frac{d}{dt}\left[\int\tau\space{d}\theta\right]$$ By the Fundamental Theorem of Calculus and the Chain Rule, (or just logic I suppose) the derivative of the RHS is the derivative of the integral multiplied by the the derivative of $\theta$.
So, $$3)\space\frac{dW}{dt} = \tau\frac{d\theta}{dt}$$ $$4)\space{P }= \tau\omega$$
Is my logic correct, but more specifically, did I appropriately take the derivative in respect to the correct variable, $\theta$? In single variable, which is the only calculus I have taken, the variable in the integrand is considered a dummy variable (I believe this is because we are taking the derivative in respect to x or t). In general, does the differential matter when taking the derivative in respect to a different variable?
It might be better to write the expression for the work done as $W(t) = \int_0^t \tau(s) \cdot \theta'(s) ds$, from which we get $P(t) = { dW(t) \over dt} = \tau(t) \cdot \theta'(t)$.