Given the operator $\Lambda=(\partial_t^2-c_1^2\partial_r^2)(\partial_t^2-c_2\partial_r^2)$ prove that the spherical mean: $$M_u(x,r,t)=\displaystyle\frac{3}{4\pi r^3}\int_{B(x,r)}u dS(r)$$ Satisfies the equation $\Lambda(M_u)=0$.
I have problems calculating the derivatives $\partial_r$, because I have to calculate til $\partial_r^4$. I know that for the first one
$$\partial_r M_u=-\displaystyle\frac{3^2}{4\pi r^4}\int_{B(x,r)}udS(r)+\displaystyle\frac{3}{4\pi r^3}\int_{\partial B(x,r)}udS(r)$$.
But I don't know how to calculate others derivatives respect to $r$. What is $\partial_r\int_{\partial B(x,r)}udS(r)$?
Any advice?
Thanks for help!