Derivative of (x^2+5x-1)/(x^2)

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Homework Help Overview

The discussion revolves around finding the first derivative of the function (x^2 + 5x - 1)/(x^2), which falls under the subject area of calculus, specifically differentiation.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute the derivative using the quotient rule but expresses uncertainty about the correctness of their result. Some participants confirm the original poster's result, while others suggest an alternative approach by simplifying the expression before differentiation.

Discussion Status

The discussion is active, with participants sharing their results and methods. There is acknowledgment of different approaches, including the quotient rule and a simplification method, but no explicit consensus on the correctness of any single method has been reached.

Contextual Notes

Participants are exploring different methods of differentiation, and there is an indication that the original poster feels their result may not be accurate. The discussion includes considerations of alternative approaches, such as the product rule, depending on the complexity of the denominator.

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Homework Statement

Find the first derivative of [...] (below)



Homework Equations



((x^2)+5x-1)/(x^2)

The Attempt at a Solution



(x^2+5x-1)/(x^2)

[tex]y' = \frac{(x^2)(2x+5-0)-((x^2)+5x+-1)(2x)}{x^4}<br /> <br /> \\ <br /> y' = \frac{2x^3+5x^2-2x^3-10x^2+2x}{x^4}<br /> <br /> \\ y' = \frac{(-5x^2)+2x}{x^4}<br /> <br /> \\ y' = \frac{-5x+2}{x^3}<br /> [/tex]

Is this correct? It feels "off."
 
Last edited:
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thats what i got... =]
 
Cool. Thank you. :)
 
However, the simplest way to do that is write
[tex]\frac{x^2+ 5x- 1}{x^2}= \frac{x^2}{x^2}+ \frac{5x}{x^2}- \frac{1}{x^2}= 1+ 5x^{-1}- x^{-2}[/tex]

The derivative of that is, of course, [itex]-5x^{-2}+ 2x^{-3}[/itex].

Can you see that that is equal to your result?
 
You can also apply the product rule if HallsofIvy's method doesn't work out (for example, if the denominator is more complicated).
 

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