How can I simplify this derivative?

AI Thread Summary
The discussion focuses on simplifying the derivative of the function h(x) = (x^2 + 3) / (2x - 1). The correct application of the quotient rule is emphasized, where h'(x) is derived as [g(x)f'(x) - f(x)g'(x)] / g(x)^2. There is clarification on not using the chain rule incorrectly, as the quotient rule already accounts for the derivatives of both the numerator and denominator. Participants express confusion about the chain rule's relevance, but it is clarified that it is not necessary in this context. Overall, the thread reinforces proper derivative techniques for rational functions.
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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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