Not Understanding Derivative Notation

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    Derivative Notation
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SUMMARY

The discussion centers on the confusion surrounding the application of Leibniz notation in classical mechanics, specifically regarding the product rule of derivatives. The user illustrates a transition from the expression "2 \dot x \ddot x" to "(v^2)'" using the product rule in reverse. The clarification provided highlights the equivalence of "2 v v'" and the understanding that both terms represent the derivative of the square of velocity. The user expresses gratitude for the explanation and acknowledges a need for better sleep to enhance focus.

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  • Understanding of Leibniz notation in calculus
  • Familiarity with the product rule of differentiation
  • Basic knowledge of classical mechanics concepts
  • Experience with derivatives and their applications in physics
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  • Study the application of Leibniz notation in various calculus problems
  • Review the product rule and its reverse applications in differentiation
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Gibby_Canes
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In my Classical Mechanics book, one of the example problems makes a step that I do not understand. The steps are in the picture in the attachment. I do not understand how the left hand side goes from what it is in the first step to what it is in the second step. I have a poor understanding of Leibniz notation, so please keep that in mind in your responses.
 

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This is just taking the product rule in reverse.

2 \dot x \ddot x = 2 v v' = (v v' + v' v) = (v^2)'
 
Wow thanks. I do not know why that had me so stumped. I should probably sleep more.

Also, I just realized this probably should've been posted elsewhere. I was doing mechanics so I just went straight to the mechanics section. Apologies to anyone reading this.
 

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