# Question on the derivative of the spherical mean

analysisintegrationmultivariable-calculus

The spherical mean of $$u$$ is defined as
\begin{align} \frac{1}{4\pi} \int_{S^2} u(\vec x + r \, \vec y) d\sigma(\vec y). \end{align}
Now this might be a dumb question, but I don't see how a $$r$$ derivative, $$\partial r$$, acts on this. In the literature it says that (if $$u$$ suff. regular)
\begin{align} \partial_r \int_{S^2} u(\vec x + r \, \vec y) d\sigma(\vec y) = \int_{S^2} \vec\nabla_x u(\vec x + r \, \vec y) \cdot \vec y \;d\sigma(\vec y). \end{align}
but why does the multivariable chain rule only produce a gradient for $$\vec x$$ instead of
\begin{align} \sum_{i=1}^3 \frac{\partial u}{\partial(x ^i + r \, y^i)} \underbrace{\frac{\partial(x^i + r \, y ^i)}{\partial r}}_{= y ^i}. \end{align}

Does it have to do with the fact that all this happens under the integral?

I think the important thing is that the spherical mean is a function of $$\vec x$$ (and $$r$$), and $$\vec y$$ is the dummy variable of integration. The choice of notation $$\nabla_{\vec x}$$ is just emphasizing that if you differentiate with respect to $$r$$, the resulting expression is still a function of $$\vec x$$ (and $$r$$). I think many authors would not even include the $$\vec x$$ in the gradient, and instead simply write $$\partial_r \int_{S^2} u(x+ry)\,d\sigma(y) = \int_{S^2}\nabla u(x+ry)\cdot y\,d\sigma(y).$$ The intermediate step is $$\int_{S^2}\partial_r\big(u(x+ry)\big)\,d\sigma(y).$$ To evaluate, set $$z(r) = x+ry$$. By the multivariable chain rule, $$\partial_r\big(u(z(r))\big) = \nabla u(z(r))\cdot z'(r) = \nabla u(x+ry)\cdot y.$$ Or with the dummy index notation, $$\partial_r\big(u(z(r))\big) = \sum_i\frac{\partial u}{\partial z_i}(z(r))\frac{dz_i}{dr}(r).$$ In this notation, $$z_i$$ is just a dummy variable.