How do I prove the following that $DA^{-1}(x)=A^{-1}$ where $A$ is a linear operator and $x \in \mathbb{R}^n$ 
How do I prove the following that $DA^{-1}(x)=A^{-1}$ where $A$ is a linear operator and $x \in \mathbb{R}^n$

I am doing the proof of the inverse function theorem and it uses the fact that $DA^{-1}(x)=A^{-1}$ but I am not sure how to show this. [Here $D$ denotes the differentiation operator]
More precisely the statement goes like this:
Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since det $f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation.
I don't understand what does differentiating a linear operator even mean?
 A: I suspect that the confusing part of this question is the fact that the derivative in this context is the total derivative. The total derivative of a function $f:\Bbb R^n \to \Bbb R^m$ at a point $x \in \Bbb R^n$ is the unique linear map $Df_x:\Bbb R^n \to \Bbb R^m$ ($Df(x)$ in the notation of your question) such that
$$
\lim_{h \to 0} \frac{\|f(x + h) - f(x) - Df_x(h)\|}{\|h\|} = 0.
$$
I find that a helpful step in making this intuitive is considering the single-variable ($m = n = 1$) case. Consider the linear function $f:\Bbb R \to \Bbb R$ given by $f(x) = ax$. Under the usual perspective on the derivative, the derivative of $f$ at any point $x \in \Bbb R$ is the number $a$. Under the total derivative perspective, the derivative of $f$ at any point $x \in \Bbb R$ is the linear transformation $Df_x(h) = ah$.
in other words, $Df_x = f$.
One point that bothered me as a student was this: doesn't this version of the derivative contradict the usual notion that the derivative of a linear function is "constant"? I would argue it doesn't: the fact that the "derivative of $f$" is constant is reflected in the fact that $Df_x(h)$ does not depend on $x$. If we consider the function $g(x) = x^2$ for the sake of contrast, we find that the total derivative of $g$ is given by $Dg_x(h) = 2xh$, corresponding to the (non-constant) "usual" derivative of $2x$.
If $f:\Bbb R^n \to \Bbb R^m$, we likewise have two perspectives on the derivative. The perspective corresponding to the "usual" approach is that the derivative of $f$ at $x$ is the Jacobian matrix $J_x$. The total-derivative perspective is that the derivative of $f$ is the linear map $Df_x(h) = J_x h$. If $f: \Bbb R^n \to \Bbb R^m$ is the linear transformation $f(x) = Mx$ for some matrix $M$, then from the first perspective we would say that the derivative of $f$ at all points is the Jacobian matrix $J_x = M$. From the total derivative perspective, the derivative is $Df_x(h) = Mh$. Again, we find that $Df_x = f$.
