I need to evaluate the following derivative in the scope of a PDE class
$\frac{d}{dt}\int_a^{x_s(t)} \rho(x,t) \text dx$
For some reason I am not able to obtain the right answer using a simple chain rule with the Fundamental theorem of Calculus. Here is what I tried (where $\frac {\partial\mu}{\partial x} =\rho$). Please note the function has a jump discontinuity at $x_s(t)$.
$\begin {align} \frac{d}{dt}\int_a^{x_s(t)} \rho(x,t) \text dx &= \frac{d}{dt} [\mu(x_s(t),t)-\mu(a,t)]\\ &= \frac{dx_s}{dt}\frac {\partial\mu}{\partial x}+\frac{\partial\mu}{\partial t}-(\frac{da}{dt}\frac {\partial\mu}{\partial x}+\frac{\partial\mu}{\partial t})\\ &= \frac{dx_s}{dt}\frac {\partial\mu}{\partial x}\\ &= \frac{dx_s}{dt}\rho \end{align}$
Which is only one term of what I should get. Any hint of where I did something wrong is appreciated, this is making me feel like I forgot calculus...
EDIT: Here is what I should get
$\frac{d}{dt}\int_a^{x_s(t)} \rho(x,t) \text dx = \int_a^{x_s(t)} \frac{\partial\rho}{\partial t} \text dx + \frac{dx_s}{dt}\rho$
The first and second $\frac{\partial \mu}{\partial t}$ are evaluated at different values $(x_s(t),t)$ and $(a,t)$. Put in these values and you will get the correct answer.