Different results with quotient rule

1. May 26, 2013

mike82140

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?

2. May 26, 2013

Hypersphere

3. May 26, 2013

mike82140

Can you show how you evaluate it? I tried both ways, but get different results.

4. May 26, 2013

Hypersphere

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,
$$\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x} = \frac{\mathrm{d}}{\mathrm{dx}} \left( x - \frac{1}{x} \right) = 1- \frac{-1}{x^2}=1+\frac{1}{x^2}$$

5. May 26, 2013

mike82140

Where did +1 come from when you were multiplying x²-1 with 1?

6. May 26, 2013

Hypersphere

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

7. May 26, 2013

mike82140

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. May 26, 2013

Hypersphere

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.

9. May 26, 2013

SteamKing

Staff Emeritus
Look at the formula for the quotient rule.

If h(x) = x, what is h'(x)?