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.

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