I'm studying in preparation for a Mathematical Analysis II examination and I came across this exercise that I cannot solve. If it's any indicator of difficulty, the exercise is Exercise 2 of 4, part $c$ and graded for 7%.
With the help of the Chain rule, prove that function $u(x, t) = f(x+ct) + g(x-ct)$ where $f,g: I \subseteq \mathbb R^2 \to \mathbb R$ are $C^2$ functions in $I$, and $c$ is a positive constant, satisfy the (wave) relation: $\dfrac{\partial^2 u}{\partial t^2} = c^2 \dfrac{\partial^2 u}{\partial x^2}$
So far I know that I should calculate partial derivatives of $u$, but I did that and yielded nothing useful. Plus, I should be using the $C^2$ clue somewhere, so I should be trying to find double derivatives throughout the whole thing.
In any case, I'm stumped. Any help would be greatly appreciated.
Applying chain rule, you have $$\begin{cases} \frac{\partial u}{\partial x}=f^\prime(x+ct)+g^\prime(x-ct)\\ \frac{\partial u}{\partial t}=cf^\prime(x+ct)-cg^\prime(x-ct) \end{cases}$$ and $$\begin{cases} \frac{\partial^2 u}{\partial x^2}=f^{\prime \prime}(x+ct)+g^{\prime \prime}(x-ct)\\ \frac{\partial^2 u}{\partial t^2}=c^2 f^{\prime \prime}(x+ct)+c^2 g^{\prime \prime}(x-ct) \end{cases}$$ Hence $$\dfrac{\partial^2 u}{\partial t^2} = c^2 \dfrac{\partial^2 u}{\partial x^2}$$ as requested.