While reading notes, I saw the following and was confused. I'm assuming that I'm wrong or I'm forgetting something obvious. I was wondering if someone could explain what I'm missing.
"If $u(x) = v(|x|)$, then $$u_{x_{i}} = v'(|x|)\frac{x_i}{|x|} , \text{ for } x\in\mathbb{R}^n\setminus\{0\}."$$
My question is why is the derivative not:
$$u_{x_{i}} = \frac{\partial v}{\partial x}\frac{\partial x}{\partial x_i} = v'(|x|)\frac{x}{|x|} x_i?$$ Where does that extra "x" go? I apologize in advance for asking such a silly question!
No, your function is a composition $u=v\circ q$, where $$u = v( q ) \atop q(x) = |x| $$
so by the chain rule the derivative is
$$u_{x_{i}} = \frac{d\,v}{d\,q}\cdot\frac{d\,q}{d\,x_i} = \frac{d\,v}{d\,\color{red}{|x|}}\cdot\frac{d\,\color{red}{|x|}}{d\,x_i} $$
and
$$\frac{d\,|x|}{d\,x_i} = \frac{d\,\sqrt{{x_i}^2 + \dots}}{d\, x_i} = \frac{2x_i}{2\sqrt{{x_i}^2 + \dots}} = \frac {x_i}{|x|}$$
Note the middle '=' above follows from the chain rule, too: $$\frac{d\,\sqrt{{x_i}^2 + \dots}}{d\, x_i} = \frac{d\,\sqrt{({x_i}^2) + \dots}}{d\, ({x_i}^2)}\cdot \frac{d\,({x_i}^2)}{d\, x_i} = \frac 1{2\sqrt{{x_i}^2 + \dots}}\cdot 2x_i$$