Review in Variable Differentiation

In summary, the conversation discusses the differentiation of a quotient using both the standard rule and the chain rule. The differentiation of D-LCos(theta) is also mentioned, with the constant "L" remaining unchanged as it is a constant. The overall topic is an explanation of an attached answer regarding an optimization problem involving minimum angle and force.
  • #1
jgreen520
9
0
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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  • #2
It uses the standard rule for differentiating a quotient: d(u/v) = (vdu - udv)/v2
 
  • #3
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]
 
  • #4
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.
 
  • #5
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
 
  • #6
"L" is a constant. What do you think should happen to it?
 

Related to Review in Variable Differentiation

1. What is variable differentiation?

Variable differentiation is a mathematical process used to find the rate of change of a dependent variable with respect to an independent variable. It involves calculating the derivative of a function, which measures the instantaneous rate of change at a specific point.

2. Why is review in variable differentiation important?

Review in variable differentiation is important because it helps to reinforce the concepts and techniques used in finding derivatives. It also allows for the identification and correction of any misunderstandings or mistakes in the application of these concepts.

3. What are the key concepts in variable differentiation?

The key concepts in variable differentiation include the definition of a derivative, the power rule, the product rule, the quotient rule, and the chain rule. It is also important to understand the concept of limits and how they relate to derivatives.

4. How is variable differentiation used in real life?

Variable differentiation has many real-life applications, particularly in fields such as physics, economics, and engineering. It is used to calculate rates of change in physical systems, such as velocity and acceleration, and to optimize functions in economics and engineering problems.

5. What are some common mistakes when applying variable differentiation?

Some common mistakes when applying variable differentiation include forgetting to use the chain rule, incorrectly applying the product and quotient rules, and making errors in algebraic simplification. It is important to carefully follow the rules and double-check calculations to avoid these mistakes.

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