Given that $$f(x,t)=\varphi (x-at)+\psi (x+at)$$ $$u=x-at$$ $$v=x+at$$
We need to prove that: $$\frac{\partial^2 f}{\partial t^2}=a^2\frac{\partial^2 f}{\partial x^2}$$
We know how to calculate the derivative of the equation with the chain rule:
$$\frac{\partial f}{\partial t}=f'_u(-a)+f'_v(a)$$ $$\frac{\partial f}{\partial x}=f'_u(1)+f'_v(1)$$
How I can calculate the second derviative ?
Thank you :)
Hint. By the chain rule, one may write $$ \frac{\partial}{\partial t}f(x,t)=\frac{\partial}{\partial t}\left(\varphi (x-at)+\psi (x+at)\right)=-a\varphi' (x-at)+a\psi' (x+at) $$ giving $$ \begin{align} \frac{\partial^2}{\partial t^2}f(x,t)&=\frac{\partial}{\partial t}\left(-a\varphi' (x-at)+a\psi' (x+at)\right) =a^2\varphi'' (x-at)+a^2\psi'' (x+at) \tag1 \end{align} $$ On the other hand, one has $$ \frac{\partial}{\partial x}f(x,t)=\frac{\partial}{\partial x}\left(\varphi (x-at)+\psi (x+at)\right)=\varphi' (x-at)+\psi' (x+at) $$ giving $$ \begin{align} \frac{\partial^2}{\partial x^2}f(x,t)&=\varphi'' (x-at)+\psi'' (x+at) \tag2 \end{align} $$ From $(1)$ and $(2)$ one gets the announced result.
