Review in Variable Differentiation

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Homework Help Overview

The discussion revolves around a variable differentiation problem within the context of optimization, specifically focusing on finding the minimum angle (theta) for the least amount of force for T_ab.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the differentiation of a quotient and explore different methods, including the standard rule and the chain rule. Questions arise regarding the treatment of constants during differentiation, particularly the constant L in the expression.

Discussion Status

The discussion is active, with participants providing various differentiation techniques and questioning the implications of constants in the differentiation process. Some guidance has been offered regarding the differentiation rules, but there is no explicit consensus on the treatment of the constant L.

Contextual Notes

Participants are working within the constraints of an optimization problem and are seeking clarification on differentiation rules as they apply to their specific scenario.

jgreen520
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Please see attached.

I was looking for an explanation of the answer I have attached. Its been a little while and was just looking for the logic behind the differentiation shown for this problem. Its basically an optimization problem where I am looking for the minimum angle (theta) for the least amount of force for T_ab.

Thanks
 

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It uses the standard rule for differentiating a quotient: d(u/v) = (vdu - udv)/v2
 
You can also do the differentiation of a quotient as the differentiation of a product using the chain rule: [itex]u/v= uv^{-1}[/itex] so that
[tex](u/v)'= (uv^{-1})'= u'v^{-1}+ u(v^{-1})'= u'v^{-1}+ u(-v^{-2}v')[/tex]
[tex]= \frac{u'}{v}+ \frac{uv'}{v^2}= \frac{u'v}{v^2}+ \frac{uv'}{v^2}= \frac{u'v+ uv'}{v^2}[/tex]
 
HallsofIvy said:
[tex]= \frac{u'}{v}+ \frac{uv'}{v^2}= \frac{u'v}{v^2}+ \frac{uv'}{v^2}= \frac{u'v+ uv'}{v^2}[/tex]
[tex]= \frac{u'}{v}- \frac{uv'}{v^2}[/tex] etc.
 
Thanks for the response.

So differentiating D-LCos(theta) the constant D goes to 0 and the -LCos(theta) goes to +Lsin(theta) ...then multiply that by sin(theta) in the denominator to get Lsin^2(theta)...

However I was curious why nothing happens to the L...in front of the Cos.

Thanks
 
"L" is a constant. What do you think should happen to it?
 

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