Different results with quotient rule

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In summary, the conversation is about finding the derivative of (x²-1)/x using the quotient rule and simplifying the expression first. The results are different and the conversation discusses why, with one person mentioning the need to include all terms in the quotient rule and the other clarifying the use of the -1 term in the formula. In the end, the correct derivative is found and the conversation concludes with appreciation for the help.
  • #1
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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  • #3
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.
 
  • #4
Alright, for a function
[tex]f(x) = \frac{g(x)}{h(x)}[/tex]
the quotient rule says
[tex]f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}[/tex]

In your case, [itex]g(x)=x^2-1[/itex] and [itex]h(x)=x[/itex]. Thus [itex]g'(x)=2x[/itex] and [itex]h'(x)=1[/itex]. Then you get
[tex]\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}[/tex]
Are you sure you included the [itex]g'(x)[/itex] term?

Simplifying first,
[tex]\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}[/tex]
 
  • #5
Where did +1 come from when you were multiplying x²-1 with 1?
 
  • #6
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.
 
  • #7
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.
 
  • #8
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 [itex]\left( x^2 -1 \right) \cdot 1[/itex], right? That is just short notation for

[tex]- \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[/tex]

Agreed? Or are you pulling my leg here?

EDIT: Ah, good. I was almost giving up for a while there.
 
  • #9
Look at the formula for the quotient rule.

If h(x) = x, what is h'(x)?
 

1. How does the quotient rule differ from the product rule in calculus?

The quotient rule is used to find the derivative of a function that is a ratio of two functions, while the product rule is used to find the derivative of a function that is a product of two functions.

2. Why do we need the quotient rule in calculus?

The quotient rule is necessary because taking the derivative of a quotient using the power rule would result in an incorrect answer. This is because the power rule only applies to products, not quotients.

3. What is the formula for the quotient rule?

The formula for the quotient rule is: (f'(x)g(x) - f(x)g'(x)) / (g(x))^2

4. Can the quotient rule be used for functions with more than two terms?

No, the quotient rule can only be applied to functions with two terms in the numerator and denominator. For functions with more than two terms, the quotient rule would need to be applied multiple times.

5. How do you know when to use the quotient rule in a calculus problem?

You should use the quotient rule when you have a function that is a ratio of two functions, and the power rule cannot be applied. Additionally, if the function contains variables in both the numerator and denominator, the quotient rule should be used.

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