Different results with quotient rule

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Homework Help Overview

The discussion revolves around finding the derivative of the function (X²-1)/X using the quotient rule and comparing it to the derivative obtained by simplifying the expression first. Participants are exploring the discrepancies in results from different methods of differentiation.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the quotient rule and simplify the expression before differentiation, leading to different results. Some participants question whether all terms were included in the application of the quotient rule. Others seek clarification on specific steps in the differentiation process.

Discussion Status

Participants are actively engaging with each other's reasoning and attempting to clarify misunderstandings. Some have provided guidance on the correct application of the quotient rule, while others are exploring the implications of simplifications made prior to differentiation.

Contextual Notes

There are indications of confusion regarding the application of the quotient rule and the distribution of negative signs in the calculations. Participants are also reflecting on their own mistakes and seeking reassurance about their understanding of the differentiation process.

mike82140
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I have been trying to figure out the derivative of (X²-1)/X. When I use the quotient rule, the result I get is 1-1/X². However, when I simplify the expression first, then take the derivative, I get 1+1/X²

Why are the results different?
 
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Hypersphere said:
I get the same answer using both methods. Are you sure you include all terms in the quotient rule? http://en.wikipedia.org/wiki/Quotient_rule

Can you show how you evaluate it? I tried both ways, but get different results.
 
Alright, for a function
f(x) = \frac{g(x)}{h(x)}
the quotient rule says
f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}

In your case, g(x)=x^2-1 and h(x)=x. Thus g'(x)=2x and h'(x)=1. Then you get
\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x}=\frac{2x\cdot x - \left( x^2 -1 \right) \cdot 1}{x^2}=\frac{2x^2-x^2+1}{x^2}=1+\frac{1}{x^2}
Are you sure you included the g'(x) term?

Simplifying first,
\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x} = \frac{\mathrm{d}}{\mathrm{dx}} \left( x - \frac{1}{x} \right) = 1- \frac{-1}{x^2}=1+\frac{1}{x^2}
 
Where did +1 come from when you were multiplying x²-1 with 1?
 
mike82140 said:
Where did +1 come from when you were multiplying x²-1 with 1?

It was also multiplied by -1, from the quotient rule formula.
 
Hypersphere said:
It was also multiplied by -1, from the quotient rule formula.

I'm sorry for asking so many questions, and this may sound stupid, but, what -1?

Edit: It appears as though I have made a mistake. The -1 is the minus part that is in front of the x²-1, so the negative, or minus, distributes itself.

Thank you for the help, I appreciate it.
 
mike82140 said:
I'm sorry for asking so many questions, and this may sound stupid, but, what -1?

You see the minus sign in front of \left( x^2 -1 \right) \cdot 1, right? That is just short notation for

- \left( x^2 -1 \right) \cdot 1=\left( -1 \right) \cdot \left( x^2 -1 \right)= (-1) \cdot x^2 + (-1) \cdot (-1) = -x^2+1

Agreed? Or are you pulling my leg here?

EDIT: Ah, good. I was almost giving up for a while there.
 
Look at the formula for the quotient rule.

If h(x) = x, what is h'(x)?
 

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