Why is this bundle homomorphism a isomorphism?

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I got trouble with an exercise:

Suppose $E$ and $E'$ are smooth vector bundles over a smooth manifold $M$,and $F:E\rightarrow{E'}$ is a bijective smooth bundle homomorphism over $M$

Prove:$F$ is a smooth bundle isomorphism i.e. $F^{-1}$ is also smooth

It seems to be rather simple,but did I miss something fundmental?

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It is enough to show that, locally, the inverse is smooth. Now locally, both $E$ and $E'$ are $U\times \mathbb R^n$, where $U\subset M$ is some open set, and $F(x,v)=(x,g(x)v)$, where $g:U\to GL(n,\mathbb R)$ is smooth. So you need to show that $g^{-1}:U\to GL(n,\mathbb R)$ is also smooth. But the latter is a composition of $g$ with the inverse map $GL(n,\mathbb R)\to GL(n,\mathbb R) $, which is smooth, because it is given by rational functions.