Calculus quotient rule problem

In summary, Homework Equations: the quotient rule g(x)f'(x) - f(x)g'(x) The Attempt at a Solution: 1/2+2x^2+3x/(x^2-5)(√x^2-5) Edit: Also, please confirm that the denominator is \sqrt{x^2} -5 as you have written, and not \sqrt{x^2 - 5}.
  • #1
thearn
27
0

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))
 
Last edited:
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  • #2
How can you simplify [itex]\sqrt{x^2}[/itex]?

Edit: Also, please confirm that the denominator is [itex]\sqrt{x^2} -5[/itex] as you have written, and not [itex]\sqrt{x^2 - 5}[/itex]. Your attempt at a solution shows that you are not always using parentheses when needed.
 
  • #3
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:
  • #4
thearn said:

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.
thearn said:

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.
 
  • #5
thearn said:
square root of x^2= x.
No, not true. ##\sqrt{x^2} = |x|##
thearn said:
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.
 
  • #6
oh ok yeah I miswrote, thanks. Yes that is valid.
 
  • #7
Mod note: To reduce confusion, I fixed the erroneous correction.

my fault yes I accidentally wrote the nine.
 
Last edited by a moderator:
  • #8
thearn said:
my fault yes I accidentally wrote the nine.

OK, so it's
[tex]y = \frac{2x-3}{\sqrt{x^2 - 5}}[/tex]
So let's identify the pieces you need for the quotient rule. What are [itex]f(x)[/itex], [itex]g(x)[/itex], [itex]f'(x)[/itex], and [itex]g'(x)[/itex]?
 
  • #9
(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
What are [itex]f'(x)[/itex] and [itex]g'(x)[/itex]?
 
  • #11
(x^2-5)^(1/2) *2 - 1/2(x^2-5)^(-1/2) *2x
-------------------------------------------------
x^2-5
 
  • #12
thearn said:
(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).
 
  • #13
jbunniii said:
What are [itex]f'(x)[/itex] and [itex]g'(x)[/itex]?

f'(x)=2 g'(x)=1/2(x^2-5)^-1/2 *2x
 
  • #14
new answer -23/2 /(x^2-5)
 
  • #15
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":

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

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:
[tex]y = \frac{2x-3}{\sqrt{x^2 - 5}}[/tex]
 
  • #16
cool, thanks for the tip, would my answer happen to be correct?
 
  • #17
If we make the fix that Mark44 pointed out, then (prior to simplifying), the quotient formula gives you
[tex]\frac{2(x^2 - 5)^{1/2} - (1/2)(2x)(2x-3)(x^2 - 5)^{-1/2}}{x^2 - 5}[/tex]
Now if you can tell us step by step how you are simplifying this, we can tell you what you are doing wrong.
 
  • #18
thearn said:
cool, thanks for the tip, would my answer happen to be correct?
Nope, unfortunately still wrong.
 
  • #19
using the chain rule shouldn't you have (1/2)(2x-3)(x^2-5)^-1/2 *2x?
 
  • #20
I edited the formula above - I accidentally left out the "2x" factor in the second term of the numerator.
 
  • #21
thearn said:
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.
 
  • #22
(2x^2 -10) -2x^2 +3/2
----------------------------------
x^2 -5*squareroot(x^2 -5)
 
  • #23
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.
 
  • #24
thearn said:
(2x^2 -10) + (-2x^2 -3/2)
----------------------------------
x^2 -5*squareroot(x^2 -5)
OK, so you multiplied the numerator and denominator by [itex](x^2 - 5)^{1/2}[/itex]. So the first term in the numerator becomes [itex]2(x^2 - 5) = 2x^2 - 10[/itex]. So far so good.

The second term becomes [itex]-(1/2)(2x)(2x-3) = -x(2x-3) = -2x^2 + 3x[/itex], 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?
 
  • #25
It's
 
  • #26
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.
 
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  • #27
thearn said:
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
[tex](2x^2 - 10) - 2x^2 + 3x[/tex]
What does this simplify to? Also, what happened to the [itex]\sqrt{x^2 - 5}[/itex] factor in your denominator?
 
  • #28
jesus christ I am sorry
 
  • #29
thearn said:
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 [itex](x^2 - 5)^{1/2}[/itex] factor in the denominator!
 
  • #30
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.
 
  • #31
thearn said:
here is my new answer i am steaming -3x-10 / x^2-5

I think [itex]3x-10[/itex] is correct for the numerator, unless I also made an algebra mistake somewhere. But you need to fix the denominator.
 
  • #32
yep its y'= 3x-10 / root(x^2-5)^3. Thanks you soooo much.
 
  • #33
thearn said:
yep its y'= 3x-10 / root(x^2-5)^3. Thanks you soooo much.

You need parentheses for the terms in the numerator. Also, mixing "root" and the exponents looks pretty weird (not wrong, though).

y' = (3x - 10)/(x2 - 5)^(3/2)
 

1. What is the quotient rule in calculus?

The quotient rule in calculus is a formula used to find the derivative of a function that is the quotient of two other functions. It states that the derivative of f(x)/g(x) is equal to (g(x)*f'(x) - f(x)*g'(x)) / (g(x))^2.

2. When do I use the quotient rule in calculus?

The quotient rule is used when you have a function that is the quotient of two other functions and you need to find its derivative. It is particularly useful when the functions involved are more complex and cannot be easily differentiated using the power rule or product rule.

3. How do I apply the quotient rule in calculus?

To apply the quotient rule, you need to identify the two functions in the quotient and their corresponding derivatives. Then, plug them into the formula (g(x)*f'(x) - f(x)*g'(x)) / (g(x))^2 and simplify the resulting expression to find the derivative of the original function.

4. What are some common mistakes when using the quotient rule in calculus?

Some common mistakes when using the quotient rule include forgetting to use the negative sign in the formula, not simplifying the resulting expression, and incorrectly identifying the two functions in the quotient. It is important to double check your work and carefully follow the steps of the quotient rule to avoid these mistakes.

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

Yes, the quotient rule can be extended to functions with more than two terms by using the chain rule. For example, if you have a function f(x) = (x^2 + 3x + 2) / (x + 1), you can rewrite it as f(x) = (x^2 + 3x + 2) * (x + 1)^-1 and then use the quotient rule to find its derivative.

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