I am reading "Calculus on Manifolds" by Michael Spivak.
In the proof of Theorem 3-13 (Change of variable), the author wrote as follows:
Since $$(g^n\circ h^{-1})'(h(a))=(g^n)'(a)\cdot [h'(a)]^{-1}=(g^n)'(a),$$ we have $D_n(g^n\circ h^{-1})(h(a))=D_ng^n(a)=1$, so that $k'(h(a))=I$.
Why did the author calculate $D_n(g^n\circ h^{-1})(h(a))$ to conclude $k'(h(a))=I$?
$$(g^n\circ h^{-1})'(h(a))=(g^n)'(a)\cdot [h'(a)]^{-1}=(g^n)'(a)＝(0,\dots,0,1)$$ because $g'(a)=I$.
From this calculation, we immediately know $k'(h(a))=I$ because $k(x)=(x^1,\dots,x^{n-1},g^n(h^{-1}(x)))$.
I cannot understand the author's intention.
Please explain why the author calculated $D_n(g^n\circ h^{-1})(h(a))$ to conclude $k'(h(a))=I$.
$D_n(g^n\circ h^{-1})(h(a))$ is the $n$th component of $(g^n\circ h^{-1})'(h(a))$.
So, $(g^n\circ h^{-1})'(h(a))$ has more information than $D_n(g^n\circ h^{-1})(h(a))$.
We need to know what the $n$th row of the matrix $k'(h(a))$ is.
I think it is not sufficient only to know the $(n,n)$ component of the matrix $k'(h(a))$ is $1$.
And the author already know what the $n$th row of the matrix $k'(h(a))$ is.
