Let $f:\mathbb{R}^+\to \mathbb{R}$ Show that $F:\mathbb{R}^2\to\mathbb{R}$, $F(x,y)=f(1+x^2+y^2)$ has a local minimum at $(0,0)$.
$f$ has positive derivatives in $\mathbb{R}^+$.
I'm struggling getting $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial y}$, because f is single variable.
Is it correct that $\frac{\partial F}{\partial x}(x,y)=f'(1+x^2+y^2)2x$ and $\frac{\partial F}{\partial y}(x,y)=f'(1+x^2+y^2)2y$ ?
Just $$x^2+y^2+1\geq1.$$ The equality occurs for $x=y=0,$ which says that we got a minimal value,
which gives that in $(0,0)$ our function has a local minimum.