# Calculus quotient rule problem

1. Oct 22, 2012

### thearn

1. The problem statement, all variables and given/known data
using the quotient rule, find the derivative of y = (2x-3)/(√(x^2-5)). Do not leave the answer in complex form.

2. Relevant equations
the quotient rule g(x)f'(x) - f(x)g'(x)
---------------------
(g(x))^2

3. The attempt at a solution

1/2+2x^2+3x/(x^2-5)(√(x^2-5))

Last edited: Oct 22, 2012
2. Oct 22, 2012

### jbunniii

How can you simplify $\sqrt{x^2}$?

Edit: Also, please confirm that the denominator is $\sqrt{x^2} -5$ as you have written, and not $\sqrt{x^2 - 5}$. Your attempt at a solution shows that you are not always using parentheses when needed.

3. Oct 22, 2012

### thearn

square root of x^2= x. The square is over the x^2 and the -5 sorry for the error. I wanted to check my answer or be guided through the problem if I am incorrect. Thanks.
Your correct in the specification of the square root. It should be the square root of x^2-5 with the square root sign over the entire expression.

Last edited: Oct 22, 2012
4. Oct 22, 2012

### Staff: Mentor

Do you mean (2x-3)/√(x^2-5)? That's different from what you wrote.
This is what you wrote:
$$\frac{1}{2} + 2x^2 + \frac{3x}{x^2 - 5} (\sqrt{x^2} - 5)$$

If that's not what you intended, please add parentheses to correct it.

5. Oct 22, 2012

### Staff: Mentor

No, not true. $\sqrt{x^2} = |x|$

6. Oct 22, 2012

### thearn

oh ok yeah I miswrote, thanks. Yes that is valid.

7. Oct 22, 2012

### thearn

Mod note: To reduce confusion, I fixed the erroneous correction.

my fault yes I accidentally wrote the nine.

Last edited by a moderator: Oct 22, 2012
8. Oct 22, 2012

### jbunniii

OK, so it's
$$y = \frac{2x-3}{\sqrt{x^2 - 5}}$$
So let's identify the pieces you need for the quotient rule. What are $f(x)$, $g(x)$, $f'(x)$, and $g'(x)$?

9. Oct 22, 2012

### thearn

(x^2-5)^1/2=g(x); f(x)=2x-3 I understand how to plug everything in I just am having problems with simplifying I suppose.

10. Oct 22, 2012

### jbunniii

What are $f'(x)$ and $g'(x)$?

11. Oct 22, 2012

### thearn

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

12. Oct 22, 2012

### Staff: Mentor

This part --> -1/2(x^2-5)^(-1/2) *2x <-- is g'(x). You are missing f(x).

13. Oct 22, 2012

### thearn

f'(x)=2 g'(x)=1/2(x^2-5)^-1/2 *2x

14. Oct 22, 2012

### thearn

15. Oct 22, 2012

### jbunniii

By the way, it's pretty easy to typeset your expressions properly using the "tex" feature, and it will make your expressions a lot easier to read. To see an example of how to do it, right-click on the following equation and select "show math as tex commands":

$$y = \frac{2x-3}{\sqrt{x^2 - 5}}$$

You can then copy/paste and modify as needed. Then add "tex" and "/tex" tags to make it work properly. For example, the equation above is specified as follows:

Code (Text):
$$y = \frac{2x-3}{\sqrt{x^2 - 5}}$$

16. Oct 22, 2012

### thearn

cool, thanks for the tip, would my answer happen to be correct?

17. Oct 22, 2012

### jbunniii

If we make the fix that Mark44 pointed out, then (prior to simplifying), the quotient formula gives you
$$\frac{2(x^2 - 5)^{1/2} - (1/2)(2x)(2x-3)(x^2 - 5)^{-1/2}}{x^2 - 5}$$
Now if you can tell us step by step how you are simplifying this, we can tell you what you are doing wrong.

18. Oct 22, 2012

### jbunniii

Nope, unfortunately still wrong.

19. Oct 22, 2012

### thearn

using the chain rule shouldn't you have (1/2)(2x-3)(x^2-5)^-1/2 *2x?

20. Oct 22, 2012

### jbunniii

I edited the formula above - I accidentally left out the "2x" factor in the second term of the numerator.