Consider a single particle (mass, $m$) that is moving with a momentum $\vec{p}$. This particle experiences a net force $\vec{F}_{net}$, which will change the particle's momentum based on the momentum principle,

$$\vec{F}_{net} = \dfrac{d\vec{p}}{dt}$$

Now, if we consider the cross product of the momentum principle with some defined lever arm (e.g., the origin of coordinates), $\vec{r}$, we can show this results in the angular momentum principle.

$$\vec{r} \times \vec{F}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$

This cross product of the lever arm and the net force is the net torque about that chosen location,

$$\vec{\tau}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$

The right hand-side of the equation can be re-written using the chain rule. This gives the difference of two terms.

$$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \dfrac{d\vec{r}}{dt} \times \vec{p}$$

The term on the far right is the cross product of the particle's velocity and momentum,

$$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \vec{v} \times \vec{p}$$

which for an object that doesn't change identity is zero.

$$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - m \underbrace{\vec{v} \times \vec{v}}_{=0}$$

And thus, we have the angular momentum principle in its derivative form,

$$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right)$$

$$\vec{\tau}_{net} = \dfrac{d\vec{L}}{dt}$$