Analysis II Basic issue applying Frechet chain rule 
Let $f$ be a $C^1$ function from $\mathbb{R^n}$ to itself. Let $T=f'(a)$ and $h(x)=T^{-1}( f(x+a)-f(a))$. Then $h'(x)=T^{-1}\circ f'(x+a)$ by the chain rule.

I can derive this result using little $o$ notation, but I clearly I've misunderstood something before because I can't see for the life of me how to get this from the chain rule.
We have $(g\circ f)'(c)=(Dg)[f(c)]\circ Df(c)$
Then in the example, set $g= T^{-1}$ and $f= f(x+a)$. We have
$$\begin{aligned} Dg&= g=T^{-1}\\Df&=f'(x+a)\end{aligned}$$
so, ignoring the constant,
$h'(x)=T^{-1}f(x+a)\circ f'(x+a)$
What went wrong? Thanks

It seems that you tried to obtained an expression for $h'(x)$ from the definition of derivative in $\mathbb{R}^n$. Here is one set of computations that you might have gone through already:
\begin{align}
h(x+k)&=T^{-1}\big(f(x+a+k)-f(x+k)\big)\\
&=T^{-1}\big(f(x+a)+f'(x+a)k+r(x+a;k)-f(x)-f'(x)-r(x;k)\big)\\
&=h(x)+\big(T^{-1} (f'(x+a)-f'(x))\big)k + R(x;k)
\end{align}
where

*

*$\lim_{k\rightarrow0}\frac{|r(x+a;k)|}{|k|}=0=\lim_{k\rightarrow0}\frac{|r(x;k)|}{|k|}$,

*$R(x;k)=T^{-1}(r(x+a;k)-r(x;k))$
($|\;|$ is the Euclidean norm in $\mathbb{R}^n$). Notice that $$|R(x;k)|\leq \|T^{-1}\|\big(|r(x+a;r)|+|r(x;k)|\big)$$
Obs: The space of matrices $\mathbb{M}_n(\mathbb{R})$ is normable. For example $\|A\|:=\sup_{|x|=1}|Ax|$ defines a norm on $\mathbb{M}_n(\mathbb{R})$ with the property that  $|Ax|\leq \|A\||x|$ for any $x\in\mathbb{R}^n$.
Hence $\lim_{k\rightarrow0}\frac{|R(x;k)}{|k|}=0$, whence
$h'(x)=T^{-1}\circ(f'(x+a)-f'(x))$ (as composition of linear operators) or $h'(x)=T^{-1}( f'(x+a)-f'(x))$ as product of matrices.
