We would also need to differentiate the solution twice if we wanted to show it satisfies the given equation. \operatorname{Let} G(s)=\int g(s) d s . \text { Which of the

following is the correct expression for } u_{t t}(x, t) ? \bigcirc \frac{c^{2}}{2}\left(f^{\prime \prime}(x+c t)+f^{\prime \prime}(x-c t)\right)+\frac{c}{2}\left(G^{\prime \prime}(x+c t)-G^{\prime \prime}(x-c t)\right) \text { О } \frac{c^{2}}{2}\left(f^{\prime \prime}(x+c t)-f^{\prime \prime}(x-c t)\right)+\frac{c}{2}\left(G^{\prime \prime}(x+c t)-G^{\prime \prime}(x-c t)\right) \text { О } \frac{c^{2}}{2}\left(f^{\prime \prime}(x+c t)+f^{\prime \prime}(x-c t)\right)+\frac{c}{2}\left(G^{\prime \prime}(s)-G^{\prime \prime}(s)\right) \bigcirc \frac{c^{2}}{2}\left(f^{\prime \prime}(x+c t)+f^{\prime \prime}(x-c t)\right)+\frac{c}{2}\left(g^{\prime \prime}(x+c t)-g^{\prime \prime}(x-c t)\right) \bigcirc \frac{c}{2}\left(G^{\prime \prime}(x+c t)-G^{\prime \prime}(x-c t)\right) \bigcirc \frac{1}{2}\left(f^{\prime \prime}(x+c t)+f^{\prime \prime}(x-c t)\right)+\frac{1}{2 c}\left(G^{\prime \prime}(x+c t)-G^{\prime \prime}(x-c t)\right)