How can I simplify this derivative?

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Homework Help Overview

The discussion revolves around simplifying the derivative of the function h(x) = (x² + 3) / (2x - 1), where participants explore the application of the quotient rule and the potential use of the chain rule in the differentiation process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the quotient rule and question the necessity of the chain rule in this context. There is confusion about the correct application of derivatives for the numerator and denominator.

Discussion Status

The discussion is active, with participants clarifying the correct use of the quotient rule. Some express uncertainty about the chain rule's relevance, while others provide guidance on the proper differentiation process.

Contextual Notes

There appears to be some misunderstanding regarding the differentiation rules, particularly the quotient rule and the chain rule, which are being examined in the context of this problem.

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Homework Statement



h(x) = (x2 + 3) / (2x - 1)
And if f = the numerator
And if g = the denominator,

h'(x) = (f/g)'
= [ (x2)' + 3' ) * (2x - 1) ] - [ ((2x)' - 1')) * (x2 + 3) ] ALL DIVIDED BY (2x - 1)2

= [2x (2x - 1) - 2 (x2 + 3) ] ALL DIVIDED BY (2x - 1)2


Would I then use the chain rule for the denominator?
 
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Why would you use the chain rule? That's the answer.
 
Pengwuino said:
Why would you use the chain rule? That's the answer.

Oh, because I thought the denominator resembled a function in a function, so I thought I could use the chain rule.
But if I did, that would be like taking the derivative of the derivative, right?
 
Sorry, I didn't see something you wrote. That's not how the quotient rule works. You don't simply take separate derivatives for the numerator and denominator. You do exactly what you already did, which was

h'(x) = {{g(x)f'(x) - f(x)g'(x)} \over g(x)^2}

That's it.
 
Pengwuino said:
Sorry, I didn't see something you wrote. That's not how the quotient rule works. You don't simply take separate derivatives for the numerator and denominator. You do exactly what you already did, which was

h'(x) = {{g(x)f'(x) - f(x)g'(x)} \over g(x)^2}

That's it.

Thanks, that makes a lot more sense.
 

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