Given two differentiable functions $f,g:V\to \mathbb{R}^n$ where $V\subset S$ is open subset of regular surface $S$.
Prove that $f+g$ with $(f+g)(x) = f(x)+ g(x)$ is still differentiable.
My attempt:for a given $x\in V$,we need to show $f+g$ is differentiable at this point,to do this we need to find one chart map $z:W\to \mathbb{R}^n$,such that $(f+g)z$ is differentiable, we have already have two different chart $a:U\to \mathbb{R}^n$ and $b:V\to \mathbb{R}^n$,such that $fa$ and $gb$ is differentiable function from $\mathbb{R}^2 \to \mathbb{R}^n$.My question is how to construct such $z$,based on $a,b$.
Hint: While in the definition only existence of one chart map is required, one usually proves next, that compositions of differentiabele functions with arbitrary chart maps is differentiable.