Show $x\mapsto x/\sqrt{1+\vert x\vert^2}$ is diffeomorphism Let $H$ be an euclidean Hilbert space, and consider $$f\colon H\to H,\quad x\mapsto\frac{x}{\sqrt{1+\vert x\vert^2}}.$$

What are $Y:=\operatorname{im}(f)$, $f^{-1}$, and $\partial f$? Show that $f\in\operatorname{Diff}^\infty(H,Y)$.

So, it should be $$\partial f(x)=\frac{\sqrt{1+\vert x\vert^2}\operatorname{id}-1/\sqrt{1+\vert x\vert^2}\langle\cdot,x\rangle x}{1+\vert x\vert^2}.$$
Is showing $(\partial f(x)h=0\Leftrightarrow h=0)$ directly, using the numerator, the right way to continue? I wasn't able to... Next, what are $Y$ and $f^{-1}$? 
 A: First we note that $f(x)$ is parallel to $x$ (since it's a product of $x$ and a non-negative real number). We use this to simplify the problem by looking at $f$ at a ray from the origin and reducing the problem to one of real function of a single variable. That is we have where $r\ge 0$:
$$f(r \overline e) = \overline e {r\over \sqrt{1+r^2}} = \overline e h(r)$$ 
So we basically need to compute the range and inverse of $h$ which is quite trivial. The range of $h$ is $[0,1)$ while it's inverse is $h^{-1}(r) = r/\sqrt{1-r^2}$. This leads to the image of $f$ is the open unit disc and the inverse of $f$ is $f^{-1}(y) = y/\sqrt{1-|y|^2}$
You actually don't have to prove that the differential is nowhere singular. What you actually need to prove is that the inverse is differentiable too which is done in very similar way.
A: $Y=\{y:|y|<1\}$ and $f^{-1} (y)=\frac y {\sqrt {1-|y|^{2}}}$. To solve the equation $y=\frac x {\sqrt {1+|x|^{2}}}$ take norm on both sides and solve for $|x|$. Then go back to the equation to see waht x is, in terms of y. Now if $\partial {f} (x)h=0$ we get $\sqrt {1+|x|^{2}} h= \frac 1 {\sqrt {1+|x|^{2}}} <h,x>x$. Hence $ ({1+|x|^{2}}) h= <h,x>x$. This gives $ ({1+|x|^{2}}) |h|=|<h,x>| |x| \leq |h| |x|^{2}$ which is possible only if $|h|=0$ or $h=0$. 
