# Elementary question in partial differentiation

Let's say we have a function of the form $f(x+vt)$ where $v$ is a constant and $x,t$ are independent variables. How is $\frac{\partial f}{\partial x} = \frac{1}{v}\frac{\partial f}{\partial t}$ equal to $f$?

If I let $u=x+vt$ then $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} = \frac{\partial f/\partial t}{\partial u/\partial t}\frac{\partial u}{\partial x}=\frac{1}{v}\frac{\partial f}{\partial t}$ but I cannot infer that $\frac{1}{v}\frac{\partial f}{\partial t} = f$ unless I assume the form of D'Alembert's Solution to be the harmonic (exponential). For the general solution I do not know how this was arrived at.

Edit: I still don't get it, as the context does not help. But I assume since it is a physics text, $f$ can be written as a Fourier series/integral of exponentials. Assuming that, the above holds.

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You are correct. Where did you see it written that you can make this inference? – Chris Taylor Oct 10 '11 at 10:33
There is a function $f:\ s\mapsto f(s)$ of one variable and a function $u:\ (x,t)\mapsto f(x+vt)$ of two variables. All you can say is that $${\partial u\over\partial x}={1\over v}{\partial u\over\partial t}=f'(x+vt)\qquad \forall x, \ \forall t\ .$$ – Christian Blatter Oct 10 '11 at 12:11
Are you by any chance of south asian descent? Your username "kuch nahi" means "nothing" in some sense in urdu/hindi – Tyler Hilton Oct 14 '11 at 2:13
@Tyler Yes (hindi). As I wrote in chat an hour ago, it is a close approximation of my progress in mathematics. – kuch nahi Oct 14 '11 at 2:41

You are correct - you can't infer that $f'(x) = f(x)$ unless $f$ is exponential, i.e. if $f(x)=A\exp(x)$.