Why is this true about the partials of this function?

Question

Consider $\rho (x, t) = f(x-t)$ where f is a fixed function. My lecturer has 'noted' that the following is true: $$\rho_t + \rho_x = 0$$

Why is that the case? ($\rho$ is the mass density) Is it physical or to do with the derivatives?

Answer 1

it's purely algebraic. you can compute partial derivatives of $\rho$ using the chain rule and see for yourself that this holds. http://en.wikipedia.org/wiki/Chain_rule

Answer 2

It is easily prooved by the chain rule:

$$\rho_t=f'(x-t)\cdot(x-t)_t=f'(x-t)\cdot(-1)$$ $$\rho_x=f'(x-t)\cdot(x-t)_x=f'(x-t)\cdot(1).$$

Add those together and you are done.