Conservation of Energy in Mechanics for Point Mass

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The discussion revolves around the simplification of an equation in atmospheric thermodynamics involving point mass mechanics. The transition from mv . dv/dt to d/dt (1/2 mv^2) is questioned, particularly how the derivative d/dt is factored out while maintaining the integrity of the velocity terms. The key point is the relationship between velocity and displacement, where v . ez equals dz/dt. The appearance of the 1/2 factor is clarified as part of the derivative of kinetic energy, which is derived from the rule for differentiating powers. Understanding these derivative manipulations is crucial for grasping the underlying physics concepts.
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Reading "Atmospheric Thermodynamics" I'm stumped almost as soon as I've started. I've probably bitten off more than I can chew and this also might even be more of a math question than a physics one but where I'm stuck is where they "simplify" from:

mv . dv/dt = -mgv . ez (where ez is a unit vector on the z-axis and the dots signify scalar multiplication by v)

to

d/dt 1/2mv . v = d/dt 1/2 mv2 = -mg dz/dtWhat I do get is that on the right we have v . ez = dz/dt, since velocity is the derivative of displacement (z) with respect to time.

What I don't get is the operation on the left by which the d/dt is just magically pulled out, leaving the v behind to work its own kind of magic on the other v. The text simply says "This equation can be simplified:" so I guess the authors presume a reader of more skill than I currently have in terms of playing with derivatives. What is it about a derivative that allows that d/dt to be stripped of its v and yanked out in front like that? And where the heck does that 1/2 appear from?
 
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It is just the derivative
\dfrac{dv^2}{dt} = 2 v \dfrac{dv}{dt}
Which comes from the rule
\dfrac{d f(t)^\alpha}{dt} = \alpha f(t)^{\alpha-1} \dfrac{df(t)}{dt}
 
Thanks Matteo, that's somewhat familiar but I'm so rusty on this stuff it's not funny.
 
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