Let $X$ be a smooth manifold, and $g:X\to \mathbb R$ be a smooth function, and everywhere positive.
Consider the map $$\varphi: T(X)\to T(X)$$ by $\varphi(x,v)=(x,g(x)v)$.
Then $\varphi$ is smooth.
Attempt: Define $S:\mathbb R\times \mathbb R^n\to \mathbb R^n$ by $S(a,\vec x)=a\vec x$ and $\triangle: \mathbb R^n\to \mathbb R^n\times \mathbb R^n$ by $\triangle(\vec x)=(\vec x,\vec x)$
Claim: $\varphi$ has smooth extention which is
if $(x,v)\in T(\mathbb R^n)$- Assume $X\subset \mathbb R^n$
$E(x,v)=(Id,S)\circ(Id,g,Id)\circ(\triangle, Id)(x,v)$
Clearly $E(x,v)=(x,g(x)v)=\varphi(x,v)$
Is it O.K. proof, and how one can prove it more elegantly and I didnot even use positiveness of the map $g$, why should this g must be positive?