I'm studying Real Analysis and my homework is to find some total derivatives using curves. For instance:
let $f:\mathbb{R}^9 \to \mathbb{R}^9$ be such that $f(M) = M^2$, then $Df_{M_0}(v)= vM_0+ M_0v, \hspace{0.2cm} \forall v \in \mathbb{R^9}$.
Clearly $f$ is differentiable. Now, choose $\gamma$ of class $C^1$ such that $\gamma:I \to \mathbb{R}^9$ and $\gamma(0)=M_0$, $\gamma'(0) = v$. Define the curve $\sigma = f \circ \gamma$.
Now, note that ${D \sigma}_0(1) = \frac {d}{dt}\sigma(t) \bigg|_{t=0}= Df_{\gamma(0)}{D\gamma}_0(1)= Df_{\gamma(0)}(v) = Df_{M_0}(v)$.
Then $D\sigma_0 (1) = D \sigma_0 = \frac{d}{dt} f \circ \gamma \bigg|_{t=0} = \frac{d}{dt}\gamma(t)^2 \bigg|_{t=0}= \gamma'(0) \gamma(0) + \gamma(0)\gamma'(0) = vM_0+M_0v.$
My questions is, how can I use that to find $Dh_0(v)$ and $Dg_0(v)$ where $h:U \subset\mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$ is given by $h(M) = M^{-1}$ and $g:\mathbb{R}^9 \to \mathbb{R}^9$ is given by $G(M) = M^t$.
I know, $h$ and $g$ are differenciables and I would like to use $\gamma$ like the example given above, but for $Dh_0$ I will have $\frac {d}{dt} \gamma^{-1}(t) \bigg|_{t=0}$ but $\gamma(t) \subset \mathbb{R}^{n^2}$, how can I solve that?
And what is $\frac{d}{dt} \gamma^t(t)$?
If $B$ is a continuous bilinear function, then $B$ is differentiable at any pair $(x, y)$ and its derivative is given by $B(x, y) \cdot (h, k) = B(x,k)+B(h,y).$ Apply this with $B(f,g)=fg$ at the point $(x, y) = (M, M)$ with $(h, k) = (v, v).$