How Does the Chain Rule Relate to Tangent Vectors in Calculus?

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SUMMARY

The discussion centers on the application of the chain rule in relation to tangent vectors in calculus, specifically addressing the equation \(\frac{dx^\nu}{d \lambda} \partial_\nu \frac{dx^\mu}{d \lambda} = \frac{d^2 x^\mu}{d \lambda^2}\). Participants explore the implications of the non-commutativity of the derivatives \(\frac{d}{d\lambda}\) and \(\partial_\mu\), leading to confusion about the calculation of the partial derivative acting on the tangent vector. The consensus emphasizes the necessity of understanding the nature of these derivatives to derive the desired result accurately.

PREREQUISITES
  • Understanding of calculus, specifically the chain rule
  • Familiarity with tangent vectors and their derivatives
  • Knowledge of partial derivatives and their properties
  • Concept of commutativity in calculus operations
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  • Study the properties of tangent vectors in differential geometry
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Homework Statement



Show that:

[tex]\frac{dx^\nu}{d \lambda} \partial_\nu \frac{dx^\mu}{d \lambda} = \frac{d^2 x^\mu}{d \lambda^2}[/tex]



The Attempt at a Solution



Well, I could simply cancel the dx^nu and get the desired result; that I do understand.
But what about actually looking at this term alone:

[tex]\partial_\nu \frac{dx^\mu}{d \lambda},[/tex]

calculating it and multiplying with dx^nu/dλ, can I get the same result? I get confused by the question: what if the partial derivative acts on the tangent vector; what happens then?


Thanks for your help!
 
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The problem is that [itex]\frac{d}{d\lambda}[/itex]and [itex]\partial_\mu[/itex] do not commute... So I'm not sure how you could calculate it without knowing what [itex]\frac{d}{d\lambda}[/itex] is.
 

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