Hello so there is a step in deriving newton's second law into the Kinetic Energy I don't quite understand it goes like this: $$\overrightarrow{F}\overrightarrow{v}=m\frac{d\overrightarrow{v}}{dt}\overrightarrow{v})$$ goes to this: $$\overrightarrow{F}\overrightarrow{v}=\frac{d}{dt}(\frac{1}{2}m\overrightarrow{v}\overrightarrow{v})$$ First I thought they just factored out $\frac{d}{dt}$, but then I couldn't explain why there is a 1/2 between the brackets.
If someone knows how to get from the first equation to the second it would be very much appreciated to tell me your thought process or whrite it out clearly
Thanks in advance
By Newton's second Law, we know that $$\sum_{i=1}^{n}\vec{F}=\vec{F}_{neta}=m\cdot \vec{a}$$ also we know that $$\frac{d\vec{x}}{dt}=\vec{v}(t)$$ $$\frac{d\vec{v}}{dt}=\vec{a}(t)$$ So, we have $$\vec{F}_{neta}\vec{v}=(m \cdot \vec{a})\vec{v}=m\frac{d\vec{v}}{dt}\vec{v}$$
By other hand, we know that $$K=\frac{1}{2}m(|\vec{v}|)^{2}$$ where $K$ is the kinetic energy and $K$ is a scalar magnitude.
Note that $$\vec{v}\cdot \vec{v}=|\vec{v}|^{2}$$where $|\vec{v}|$ denote the magnitude of the vector $\vec{v}$.
Also, note that $$\frac{d}{dt}\left( \frac{1}{2}m \vec{v}\cdot \vec{v}\right)=\frac{1}{2}m \frac{d}{dt}(\vec{v}\cdot \vec{v})=m\frac{d|\vec{v}|}{dt}$$