Answer
We show that $\frac{{\partial u}}{{\partial t}} + c\frac{{\partial u}}{{\partial x}} = 0$
Work Step by Step
We have $u\left( {x,t} \right) = f\left( {x - ct} \right)$.
Let $r\left( {x,t} \right) = x - ct$.
So, $u\left( {x,t} \right) = f\left( {r\left( {x,t} \right)} \right)$.
The partial derivatives are
$\frac{{\partial u}}{{\partial x}} = \frac{{df}}{{dr}}\frac{{\partial r}}{{\partial x}} = \frac{{df}}{{dr}}$
$\frac{{\partial u}}{{\partial t}} = \frac{{df}}{{dr}}\frac{{\partial r}}{{\partial t}} = - c\frac{{df}}{{dr}}$
So,
$\frac{{\partial u}}{{\partial t}} + c\frac{{\partial u}}{{\partial x}} = - c\frac{{df}}{{dr}} + c\frac{{df}}{{dr}} = 0$
Hence,
$\frac{{\partial u}}{{\partial t}} + c\frac{{\partial u}}{{\partial x}} = 0$
