Problem: Let $\xi=x-ct$ and $\eta=x+ct.$ Show that the initial value problem
$$u_{tt}=c^2u_{xx}, \quad u(x,0)=f(x), \quad u_t(x,0)=g(x) \tag1$$
satisfies d'Alembert's formula
$$u(x,t)=\frac{1}{2}[f(x-ct)+f(x+ct)]+\frac{1}{2c}\int\limits_{x-ct}^{x+ct}g(y) \ dy. \tag{2}$$
I think I'm on the right track here but I just can't spot my mistake. Can anyone see where my arithmetic has gone south?
We can rewrite $(2)$ as
$$\frac{1}{2}\left[f(\xi)+f(\eta)+\frac{1}{c}(G(\eta)-G(\xi))\right], \tag3$$
where $G$ is the primitive of $g$. Now,
$$u_t = \frac{\partial u}{\partial\xi}\frac{\partial \xi}{\partial t}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial t}=\frac{g(\xi)}{2}-\frac{cf'(\xi)}{4}+\frac{g(\eta)}{2}+\frac{cf'(\eta)}{4},\tag4$$
$$u_x=\frac{\partial u}{\partial\xi}\frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x} = \frac{f'(\xi)}{4}-\frac{g(\xi)}{2c}+\frac{f'(\eta)}{4}+\frac{g(\eta)}{2c}.\tag5$$
We can also rewrite $(1)$ as
$$u_{tt}=c^2u_{xx}=0\Leftrightarrow\left(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\right)u=0. \tag6$$
Pluging in $(4)$ & $(5)$ we get
$$\left(\frac{g(\xi)}{2}-\frac{cf'(\xi)}{4}+\frac{g(\eta)}{2}+\frac{cf'(\eta)}{4}-c\left[\frac{f'(\xi)}{4}-\frac{g(\xi)}{2c}+\frac{f'(\eta)}{4}+\frac{g(\eta)}{2c}\right]\right)\times \times\left(\frac{g(\xi)}{2}-\frac{cf'(\xi)}{4}+\frac{g(\eta)}{2}+\frac{cf'(\eta)}{4}+c\left[\frac{f'(\xi)}{4}-\frac{g(\xi)}{2c}+\frac{f'(\eta)}{4}+\frac{g(\eta)}{2c}\right]\right) \tag7$$
After simplifying and canceling terms I get
$$\left(g(\xi)-c\frac{f'(\xi)}{2}\right)\left(g(\eta)+c\frac{f'(\eta)}{2}\right)\neq0.\tag8$$
Any help is appreciated.