derivatives of Kinetic Energy I read that derivative of Kinetic Energy function = $F.v$ while I got $mv$ when I differentiated it with respect to velocity. 
The way I did it is: 
$\frac{dK}{dv} = \frac{1}{2} m . \frac{d}{dv} v^2$ 
So I assumed that the mass is fixed and I differentiated the squared velocity by taking the (2) down and subtracting (1) from the exponent, which gave me $2v$ for $\frac{d}{dv} v^2$
Would you clarify the correct derivative of Kinetic Energy as well as the $F$ variable? And what does the derivative in this case represent?

The derivate of kinetic energy respect to the time $t$ is $Fv$:
$$K'=mvv'=mva=Fv$$
In general $v$ depends by time so the total derivative of $K$ is $Fv$, i.d. the instantaneous power.
