Show the following rule for differentiable curves in $\mathbb{R}^3$:
$$\frac{d}{dt}\left \{\overrightarrow{\sigma}(t) \cdot \left [\overrightarrow{\rho}(t)\times \overrightarrow{\tau}(t)\right ]\right \}=\frac{d\overrightarrow{\sigma}}{dt}\cdot \left [\overrightarrow{\rho(t)}\times \overrightarrow{\tau}(t)\right ]+\overrightarrow{\sigma}(t)\cdot \left [\frac{d\overrightarrow{\rho}}{dt}\times \overrightarrow{\tau}(t )\right ]+\overrightarrow{\sigma}(t)\cdot \left [\overrightarrow{\rho}(t)\times \frac{\overrightarrow{\tau}}{dt}\right ]$$
Could you give me some hints how we could show this??
You will need the product rule for dot products: $$ \frac{d}{dt}\left\{\alpha(t)\cdot\beta(t)\right\} = \left\{\frac{d}{dt}\alpha(t)\cdot\beta(t)\right\} + \left\{\alpha(t)\cdot\frac{d}{dt}\beta(t)\right\}. $$ You will also need the product rule for cross products: $$ \frac{d}{dt}\left\{\alpha(t)\times\beta(t)\right\} = \left\{\frac{d}{dt}\alpha(t)\times\beta(t)\right\} + \left\{\alpha(t)\times\frac{d}{dt}\beta(t)\right\}. $$ Apply these to $\alpha = \sigma$, $\beta = \rho\times\tau$.