I worked on one problem I stuck because I think I have a problem about applying chain rule at function with more than one variable. I will write what I try to do, but please explain my step by step so that can I understand what differental operators in chain rule do. $$f(x)=arctg(||x||^2)*x$$ Find a $Df(x)(a)$ if $x = (1, 1, ..., 1)$ and $ a = (1, 2, ..., n)$. My work: It is easy to see that we will need chain rule and product rule to solve this problem. First of all let's look at $arctg(||x||^2)$.
This function we can write as $arctg\circ n$. Where $n$ is $n(x)=||x||^2$. I write in that way because we know $||x||^2=<x,x>$ and we can easily compute derivate of n(product rule). $Dn(x)(h)=2<x,h>$. And for $arctg(x)$ we know that derivative of $arctg(x)$ is $\frac{1}{1+x^2}$.
But I have problem in using chain rule...
$f = \phi \circ (\arctan \circ \|\cdot\|^2) = \phi \circ g$, where $\phi: t \in \Bbb R \mapsto \phi(t) = tx \in \Bbb R^n$.
We have $Dg(x) = D(\arctan)(\|x\|^2) D(\|\cdot\|^2)(x)$, i.e. for each $h$,
$$Dg(x)h = \frac{2\langle x,h\rangle}{1 + \|x\|^4} $$
Also, as $\phi$ is linear and continuous, $D\phi(t) = \phi$. Thus: $Df(x) = D\phi(g(x)) \circ Dg(x) = \phi \circ Dg(x)$. In other words, for each $h$:
$$Df(x)h = \phi(Dg(x)h) = \frac{2\langle x,h\rangle}{1+\|x\|^4} x$$