How to compute $u_x(\sqrt {(x^2+y^2)})=$? This is in reference to this question 
$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$
If $u$ is a function of single variable $x$ and if I want to differentiate $u(\sqrt {(x^2+y^2)}$ with respect to $x$
then what will be $u_x(\sqrt {(x^2+y^2)})=$?
It is given in the question that linked above
that $u_x(\sqrt {(x^2+y^2)})=u_x\times \frac{x}{\sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false
What is the correct way to solve it?
Any help.

What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=u\left(\sqrt{x^2+y^2}\right)=x^2+y^2+\sqrt{x^2+y^2}.$$
Then
$$\frac\partial{\partial x}f(x,y)=2x+\frac{x}{\sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'\left(\sqrt{x^2+y^2}\right)\frac\partial{\partial x}\sqrt{x^2+y^2}
=\left(2\sqrt{x^2+y^2}+1\right)\frac{x}{\sqrt{x^2+y^2}}
=2x+\frac{x}{\sqrt{x^2+y^2}}.$$
I don't see any problem....
