I'm working through this PDE textbook and am trying to understand where I'm going wrong in reproducing a computation of a second partial derivative $u_{x_i x_i}$ in the text (for reference, $\S$2.2.1.a, p.21).
We start with $u(x) = v(r)$, where $r(x) = \|x\|_2 = (x_1^2+\cdots+x_n^2)^{-1/2}$, then compute $$\frac{\partial r}{\partial x_i} = \frac{1}{2}(x_1^2+\cdots+x_n^2)^{-1/2} 2x_i = \frac{x_i}{r}$$ Fine, so far so good. The first derivative $u_{x_i}$ follows easily: $$u_{x_i} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x_i} = u_r \frac{x_i}{r} = v'(r)\frac{x_i}{r}$$ The trouble (for me) comes in computing the second derivative. The text states $$u_{x_i x_i} = v''(r)\frac{x_i^2}{r^2} + v'(r)\left(\frac{1}{r}- \frac{x_i^2}{r^3}\right)$$
When I work through it myself, I find \begin{align*} u_{x_i x_i} = \frac{\partial u_{x_i}}{\partial r}\frac{\partial r}{\partial x_i} &= \frac{\partial}{\partial r}\left[ v'(r)\frac{x_i}{r}\right]\cdot\frac{x_i}{r} \\ &= \left( v''(r)\frac{x_i}{r} + v'(r)\left(-\frac{x_i}{r^2}\right)\right)\frac{x_i}{r} \\ &= v''(r)\frac{x_i^2}{r^2} - v'(r)\frac{x_i^2}{r^3} \end{align*}
It seems I must be missing (or forgetting) something. In order to arrive at the result in the text, it would appear that we should have $$\frac{\partial}{\partial r}\left[ \frac{x_i}{r} \right] = \frac{1}{x_i}- \frac{x_i}{r^2},$$ but why would that be?
Begin by differentiating the product, $$ \frac{\partial}{\partial x_i} \left( \frac{v'(r)}{r} \cdot x_i \right) = \frac{\partial}{\partial x_i} \left( \frac{v'(r)}{r} \right) \cdot x_i + \frac{v'(r)}{r} \cdot 1= \dots, $$ then use the chain rule. What you forgot is that $x_i$ depends on $r$.