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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!

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Why is there a $y$ in your title? Is $x$ actually a vector? – anon Apr 12 '12 at 21:33
up vote 1 down vote accepted

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,

$$\frac{du}{dt}=\frac{\partial u}{\partial x} \frac{\partial x}{\partial t}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial t}=e^{-rt}\cdot x'(t)+(-rxe^{-rt}). $$

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