Let $F : M \to \mathbb{R}^{m\times k}$ (where $\mathbb{R}^{m\times k}$ are $m\times k$ matrices with real coefficients) and $G : M \to \mathbb{R}^{k\times n}$.
Define $H : M \to \mathbb{R}^{m\times n}$ by $H(x)=F(x)G(x)$. I want to show that $dH_x(v) = dF_x(v)G(x)+F(x)dG_x(v)$ but I am not sure how to start.
Take $\gamma:(-\epsilon,\epsilon)\rightarrow M$ such that $\gamma(0)=x$ and $\gamma'(0)=v$ then the chain rule said that $$ dH_x(v)=(H\circ \gamma)'(0) $$
but $H\circ \gamma$ it's now a curve on $\mathbb{R}^{m\times n}$ given by $$ F(\gamma(t))G(\gamma(t)) $$. if $F=(F_{ij})$ and $G=(G_{ij})$ then $$ (H\circ \gamma(t))_{ij}=\sum_{k}F_{ik}(\gamma(t))G_{kj}(\gamma(t)) $$ then $$ \frac{d}{dt}(H\circ \gamma(t))_{ij}=\sum_{k}\frac{d}{dt}(F_{ik}(\gamma(t)))G_{kj}(\gamma(t))+\sum_{k}F_{ik}(\gamma(t))\frac{d}{dt}(G_{kj}(\gamma(t))) $$
by the chain rule on $\mathbb{R}$, thus $$ \frac{d}{dt}|_{t=0}(H\circ \gamma(t))_{ij}=\sum_k d(F_{ik})_x(v)G_{kj}(x)+\sum_k F_{ik}(x)d(G_{kj})_x(v) $$
thus $$ dH_x(v)=dF_x(v)G(x)+F(x)dG_x(v) $$