Given $u(t,x) = xe^{-rt}$ and $x(t)$
what is $\frac{\partial u}{\partial t} ={}$?
I would like to understand the way it works. Thanks!
Given $u(t,x) = xe^{-rt}$ and $x(t)$
what is $\frac{\partial u}{\partial t} ={}$?
I would like to understand the way it works. Thanks!
The partial derivative is the derivative of $u$ with respect to one of its arguments alone, as if all other arguments it depends on are fixed. So here $\partial_t u(x,t)=\partial_t (xe^{-rt})=-rxe^{-rt}=-ru$. The total derivative on the other hand lets everything vary concurrently and thus obeys the chain rule,
$\frac{du}{dt}=\frac{\partial u}{\partial x} \frac{\partial x}{\partial t}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial t}=e^{-rt}\cdot x'(t)+(-rxe^{-rt}). $