Understanding Derivatives: The Role of the Chain Rule Explained | Homework Help

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Homework Help Overview

The discussion revolves around understanding the application of the chain rule in calculus, specifically in relation to derivatives involving the arcsecant function. Participants are examining a problem from a homework assignment that involves differentiating a function related to arcsec.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster questions why the derivative includes an x instead of a 2x, suggesting confusion about the chain rule's application. Other participants discuss simplification and the role of the exponent when differentiating, while also clarifying terminology related to the arcsec function.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the derivative of the arcsec function. There is recognition of mistakes made in previous attempts, and a collaborative effort to clarify misunderstandings about the chain rule and its implications in this context.

Contextual Notes

Some participants express uncertainty stemming from their previous calculus experiences, indicating a potential gap in foundational understanding that may affect their current problem-solving abilities.

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Homework Statement


I attached a problem from my homework.

My only issue is that I don't know why it is x arc sec and not 2x arc sec etc...shouldn't there be a 2x from the chain rule from the derivative inside the square root? I cannot understand for my life why there is an x and not a 2x.

I kept getting this wrong until I googled it and someone in yahoo answers gave a response without a 2, so i tried it and my online homework said it was right. But don't know why!


Homework Equations





The Attempt at a Solution

 

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Well, I didn't get the same result, maybe I have to simplify it, but the 2 factor goes away because the 1/2 exponent "comes down" when you derive sqrt (x^2 - 1). Is that what you are talking about? And why are you talking about arcsec?
 
Oh dear. That was a silly mistake! I definitely didn't bring down the 1/2 when I took the derivative. ::slaps forehead:: It's been a while since I took calc 1 so I am making a lot of really silly mistakes now in Calc 2!

I couldn't see for my life what I was missing!

Oh and I was saying arcsec because i thought sec^-1x was more confusing to read.

Thank you for your help!
 
1/sec = cos btw
 
Well

[tex](\sec^{-1} x)' = \frac{1}{x\sqrt{x^2 -1}}[/tex]

so that explains the first term. As for the second, there's no 2, because of the chain rule.
 

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