I'm working on a question from a past exam paper (Mathematics BSc, second-year module in differential equations, unpublished). Here's a paraphrase of the question:
Consider the first-order PDE $$\frac{\partial u}{\partial t}+V\frac{\partial u}{\partial x}=0,\tag{1}$$ where $V>0$ is a constant. The general solution [found in part of the question] is $u(x,t)=g(x-Vt)$, where $g$ is an arbitrary function.
Under a certain condition [proved as part of the question; I don't think it's needed here], $(1)$ is satisfied by $$u(x,t)=\frac{\partial q}{\partial t}-V\frac{\partial q}{\partial x},\tag{2}$$ where $q(x,t)$ is another function.
The function $q$ satisfies the initial conditions $$q(x,0)=\frac{1}{1+x^4},\;\;\frac{\partial q}{\partial t}(x,0)=0.$$
Find the solution of $(1)$ for the $u$ given in $(2$) [I think "subject to these initial conditions" is implied].
For part of the solution, the answer booklet has,
At time $t=0$, $$u(x,0)=-V\frac{\partial}{\partial x}\left(\frac{1}{1+x^4}\right)=\frac{4Vx^3}{1+x^4}=g(x).$$
This seems to use $\frac{\partial}{\partial x}(q(x,0))$ where I would expect to see $\frac{\partial q}{\partial x}(x,0)$. In general, these aren't the same, right? I also get $-V\frac{\partial}{\partial x}\left(\frac{1}{1+x^4}\right)=\frac{4Vx^3}{(1+x^4)^2}$. Could someone help me understand what's gone wrong?