Calculus quotient rule problem

[thearn]

Homework Statement

using the quotient rule, find the derivative of y = (2x-3)/(√(x^2-5)). Do not leave the answer in complex form.

Homework Equations

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

The Attempt at a Solution

1/2+2x^2+3x/(x^2-5)(√(x^2-5))

 

[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.

 

[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.

 

[Mark44]

Homework Statement

using the quotient rule, find the derivative of y = (2x-3)/(√x^2-5). Do not leave the answer in complex form.

Do you mean (2x-3)/√(x^2-5)? That's different from what you wrote.

Homework Equations

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

The Attempt at a Solution

1/2+2x^2+3x/(x^2-5)(√x^2-5)

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.

 

[Mark44]

square root of x^2= x.

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

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.

 

[thearn]

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

 

[thearn]

Mod note: To reduce confusion, I fixed the erroneous correction.[/color]

my fault yes I accidentally wrote the nine.

 

[jbunniii]

my fault yes I accidentally wrote the nine.

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 $g'(x)$?

 

[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.

 

[jbunniii]

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

 

[thearn]

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

 

[Mark44]

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

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

 

[thearn]

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

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

 

[thearn]

new answer -23/2 /(x^2-5)

 

[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:

[tex]y = \frac{2x-3}{\sqrt{x^2 - 5}}[/tex]

 

[thearn]

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

 

[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.

 

[jbunniii]

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

Nope, unfortunately still wrong.

 

[thearn]

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

 

[jbunniii]

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

 

[jbunniii]

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

Yes, fixed it. Thanks. So what's a good first step for simplifying the fraction? Hint: try to get rid of the nasty 1/2 and -1/2 powers in the numerator.

 

[thearn]

(2x^2 -10) -2x^2 +3/2
----------------------------------
x^2 -5*squareroot(x^2 -5)

 

[thearn]

I multiplied both the numerator and the denominator by the square root of x squared minus 5.

I have to get use to this equation writing thing. It's quite hard otherwise.

 

[jbunniii]

(2x^2 -10) + (-2x^2 -3/2)
----------------------------------
x^2 -5*squareroot(x^2 -5)

OK, so you multiplied the numerator and denominator by $(x^2 - 5)^{1/2}$. So the first term in the numerator becomes $2(x^2 - 5) = 2x^2 - 10$. So far so good.

The second term becomes $-(1/2)(2x)(2x-3) = -x(2x-3) = -2x^2 + 3x$, which isn't quite the same as what you have. I think you are just making simple algebra mistakes. Are you writing this down on paper or trying to do it in your head?

 

[thearn]

It's

 

[thearn]

oh alright, 3x-10/ x^2-5 I was just made a whole bunch of little errors. thanks. Yeah I was writing it out but I guess I am just a little tired.

 

[jbunniii]

yeah i corrected it. My fault. answer -2x^2 -17/2 / (x^2-5)

That's still not what I get. Adding my first and second term, the numerator should be
$$(2x^2 - 10) - 2x^2 + 3x$$
What does this simplify to? Also, what happened to the $\sqrt{x^2 - 5}$ factor in your denominator?

 

[thearn]

jesus christ I am sorry

 

[jbunniii]

oh alright, 3x-10/ x^2-5 I was just made a whole bunch of little errors. thanks. Yeah I was writing it out but I guess I am just a little tired.

OK, that's what I get too. No worries, we all have days where can barely calculate 2+2.

[edit] Wait, you're still missing a $(x^2 - 5)^{1/2}$ factor in the denominator!

 

[thearn]

jesussss... I'm just being so dumb leme do this problem fifteen times before I write anything else so I can cease to waste your time.