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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 $||x||^2=$ and we can easily compute derivate of n(product rule). $Dn(x)(h)=2$. 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...

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    You should tell us the domain of f and where it maps to. Also, try \arctan in between $ signs. – 2017-01-10
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    Domain of f sholud be all real numbers. – 2017-01-10
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    Oh come on, you have $n$-vectors there. – 2017-01-10
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    Sorry, domain is R^n – 2017-01-10

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$f = \phi \circ (\arctan \circ \|\cdot\|^2) = \phi \circ g$, 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 $h$:

$$Df(x)h = \phi(Dg(x)h) = \frac{2\langle x,h\rangle}{1+\|x\|^4} x$$