Proving differentiability and calculating the differential of $A \mapsto A^2$ 
Let $$\begin{aligned} f : M_n(\Bbb R) &\to M_n(\Bbb R)\\ A  &\mapsto A^2 \end{aligned}$$ Prove that $f$ is differentiable and calculate its differential.


Any hints on how to prove that? I don't think $f$ is linear so any theorems I can use?
 A: Fix $A\in M_n(\mathbb{R})$. As a function on the space of matrices
$$
f(A+H)=(A+H)^2 = A^2+ AH + HA +H^2
$$
The transformation $T:M_n(\mathbb{R})\rightarrow M_n(\mathbb{R})$ given by
$$T(H)=AH+HA$$
is linear (consider $cH_1+H_2$, $c\in\mathbb{R}$ and $H_j\in M_n(\mathbb{R})$.
For the norm $\|M\|_{2,\infty}=\sup_{|x|_2=1}|Mx|_2$ in $M_n(\mathbb{R})$ for example, one has that $\|H^2\|\leq \|H\|\|H\|$ and so
$$
\frac{\|H^2\|}{\|H\|}\leq \frac{\|H\|\|H\|}{\|H\|}\xrightarrow{\|H\|\rightarrow0}0
$$
From all this, it follows that $f$ is differentiable at any $A\in M_n(\mathbb{R})$ and $Df(A)H=AH+HA$.
Remark:
There are other many norms on $M_n(\mathbb{R})$ on which $\|AB\|\leq\|A\|\|B\|$ for any $A,B\in M_n(\mathbb{R})$. We take one for convenience. Any other norm, say $N$, being equivalent to $\|\,\|_{2,\infty}$ would also give the key part:
$\lim_{N(H)\rightarrow0}\frac{N(H^2)}{N(H)}=0$.
A: I would suggest that you prove that, if $f:U \times V \to W$ is a bilinear map, then $f$ is differentiable and $df_{(u_0,v_0)}(u,v)= f(u_0,v)+f(u,v_0)$. Here $U,V,W$ are finite-dimensional real vector spaces.
The map you are looking at is the composition of the (linear) diagonal inclusion map $A \mapsto (A,A) : M_n(\mathbb{R})\to M_n(\mathbb{R})\times M_n(\mathbb{R})$ with the (bilinear) multiplication map $(A,B) \mapsto AB:M_n(\mathbb{R}) \times M_n(\mathbb{R}) \to M_n(\mathbb{R})$.
