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I am currently in Mechanics I and both my professor and my book have derived the work principle in this way and I even asked about its derivation during class, but it has me puzzled.

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I don't understand how they justify $$d(v.v)/dt = 2v.v' .$$ I'm assuming they take the $m$ out as a constant and but if they are taking the derivative with respect to time how does the 2 come in. Further more where does the half come from in 4.1.4. I recognize the$ (1/2) mv,$ but why wouldn't it be $(1/2) mv^2?$ I would appreciate any insight, I'm at a loss.

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    $\begingroup$ Do you know how to take the derivative of a product? Do that and you will find the 2. It's not a physics thing; it's a math thing. $\endgroup$
    – Bill N
    Commented Oct 15, 2021 at 1:17
  • $\begingroup$ You realize that $v^2$ and $\bf v\cdot v$ are the same thing yes? $\endgroup$
    – joseph h
    Commented Oct 15, 2021 at 1:59
  • $\begingroup$ Indeed, that is true that the dot product of v and v would be 2v. and indeed the derivative of 2v would be 2. it was the v dotted with dv/dt (the second v is a v with a dot on top indicating derivative that confused me) that's why it confused me. $\endgroup$
    – HappyHiggs
    Commented Oct 15, 2021 at 2:17
  • $\begingroup$ Note that the derivative of 2v with respect to v is 2, but the derivative with respect to time is $2\dot v$. $\endgroup$
    – joseph h
    Commented Oct 15, 2021 at 2:33
  • $\begingroup$ Thankyou, I think that's what I was missing that the 2v(dot) represents the derivative with respect to time. $\endgroup$
    – HappyHiggs
    Commented Oct 15, 2021 at 22:33

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