If $f:\mathbb{R}^{m\times m}\to\mathbb{R}^{m\times m}$ is given by $f(M)=M^5$, find $D^2f(X)(H,K)$ for $X,H,K\in\mathbb{R}^{m\times m}$.
I know how to find $D^1f(X)(H)$ by doing: \begin{align*} D^1f(X)(H)&=\lim_{t\to 0}\frac{f(X+tH)-f(X)}{t}\\ &=\lim_{t\to 0}\frac{(X+tH)^5-X^5}{t}\\ &=X^4H+X^3HX+X^2HX^2+XHX^3+HX^4 \end{align*}
But when I try to do a similar computation for $D^1f$, namely $\lim_{t\to 0}\frac{D^1f(X+tH)-D^1f(X)}{t}$, I get confused with the meaning of the limit.
I guess my confusion is that $D^1f$ is a function from $\mathbb{R}^{m\times m}$ to $\mathcal{L}(\mathbb{R}^{m\times m})$ (linear maps from $\mathbb{R}^{m\times m}$ to itself), and I don't know how to deal with $\mathcal{L}(\mathbb{R}^{m\times m})$ when it comes to derivation.
As a simpler case, let's try $f(M)=M^3$. In this case,
\begin{align*} D^1f(X)(H)&=\lim_{t\rightarrow 0}\frac{f(X+tH)-f(X)}{t}\\ &=\lim_{t\rightarrow 0}\frac{X^3+tX^2H+tXHX+tHX^2+t^2XH^2+t^2HXH+t^2H^2X+t^3H^3-X^3}{t}\\ &=\lim_{t\rightarrow 0}(X^2H+XHX+HX^2+tXH^2+tHXH+t2H^2X+t^2H^3)\\ &=X^2H+XHX+HX^2. \end{align*}
Now, using the formula from above, we might consider \begin{align*} \lim_{t\rightarrow 0}&\frac{D^1f(X+tK)(H)-D^1f(X)(H)}{t}\\ &=\lim_{t\rightarrow 0}\frac{(X+tK)^2H+(X+tK)H(X+tK)+H(X+tK)^2-(X^2H+XHX+HX^2)}{t}\\ &=\lim_{t\rightarrow 0}\left(\frac{X^2H+tXKH+tKXH+t^2K^2H}{t}+\frac{XHX+tXHK+tKHX+t^2KHK}{t}+\frac{HX^2+tHXK+tHKX+t^2HK^2}{t}-\frac{X^2H+XHX+HX^2}{t}\right)\\ &=\lim_{t\rightarrow 0}(XKH+KXH+tK^2H+XHK+KHX+tKHK+HXK+HKX+tHK^2)\\ &=XKH+KXH+XHK+KHX+HXK+HKX \end{align*}
If you follow the same procedure, but reversing the roles of $H$ and $K$, you'll get the same answer because the answer is symmetric in the roles of $H$ and $K$. If I try to type it out, I'm sure that I'm going to make an error. But the final answer is that both limits above give the same answer.