The equations I have to use are:
$$\underline r = (x_{1},x_{2},x_{3})$$ and $$ r = \vert \underline r \vert$$
Can someone explain how the following equation
$$\frac{1}{2}(x_{j}x_{j})^{-1/2}\frac{\partial}{\partial x_{i}}(x_{j}x_{j})$$
Simplifies into the equation below.
$$ = \frac{1}{2r}2x_{j}\frac{\partial x_{j}}{\partial x_{i}}$$
I'm not 100% certain this is correct but it was the material given to me.
Using Einstein's notation
$$\frac{1}{2}(x_{j}x_{j})^{-1/2}\frac{\partial}{\partial x_{j}}(x_{j}x_{j})$$
Here a summation is performed over the free indexes and is equivalent to
$$ \frac{1}{2}\frac{2x_j}{\sqrt{x_1^2+x_2^2+x_3^2}} = \frac{\partial}{\partial x_j}\sqrt{x_1^2+x_2^2+x_3^2} $$
then as $r^2 = x_1^2+x_2^2+x_3^2$
$$ \frac{1}{2}(x_{i}x_{i})^{-1/2}\frac{\partial}{\partial x_{j}}(x_{i}x_{i}) = \frac{1}{2}\frac{2x_j}{\vert r\vert}\delta_i^j $$
Here $\delta _i^j = \frac{\partial x_j}{\partial x_i}$ is the Kronecker delta function.