What is the Derivative of 1/(x√(x^2-1))?

  • Context: Undergrad 
  • Thread starter Thread starter Naeem
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around finding the derivative of the function \( \frac{1}{x\sqrt{x^2-1}} \) and exploring methods for integrating it. Participants are engaged in a mix of derivative and antiderivative calculations, with some focusing on substitution techniques and others questioning the use of hyperbolic functions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents an initial approach to differentiate \( \frac{1}{x\sqrt{x^2-1}} \) but expresses uncertainty about its correctness.
  • Another participant suggests using a hyperbolic substitution involving \( \cosh u \) for integration.
  • A different participant questions the prevalence of hyperbolic functions in education, noting they were not covered in their high school curriculum.
  • There is a discussion about simplifying the expression before integration, with a focus on substitution methods.
  • One participant challenges the correctness of the initial approach and suggests an alternative substitution involving \( \sec u \).
  • Further calculations are presented, including the relationship between \( x \) and \( u \) using the secant function.
  • Another participant applies initial conditions to derive a specific expression for \( y \) based on the value of \( x \). They inquire about the possibility of rewriting the expression in a different form.
  • A later reply notes the multivalued nature of the secant function and provides a specific value for \( \sec^{-1}(2) \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the initial approach, with some challenging it and others proposing alternative methods. The discussion remains unresolved regarding the best approach to take.

Contextual Notes

There are limitations in the discussion regarding the assumptions made in the initial approach, the dependence on specific substitutions, and the unresolved steps in the differentiation and integration processes.

Naeem
Messages
193
Reaction score
0
Hi,

Can anyone help me here!

dy/dx = 1/(x.sqrt(x^2-1)) , x>0 Note y = pi when x = 2

Here is what I did:

Let u = x^2 - 1
x^2 = u + 1

Substituting this in the above yields:

dy/dx = 1/x.sqrt(u)

Differentiating: u = x^2-1

du = 2x.dx

x = du/2.dx

Then,

dy/dx= 2/sqrt(u).du

I don't know if this is rite.
Plz help me!
 
Physics news on Phys.org
What are u trying to do?Find the antiderivative of [itex]\frac{1}{x\sqrt{x^{2}-1}} [/tex] ...?If so,then use a substitution involving [itex]\cosh u[/itex]...<br /> <br /> Daniel.[/itex]
 
Why is it that everyone has been taught about hyperbolic trig functions when we skipped it in high school and calc 1-3?
 
I was trying to simplify the Right Hand side, before, I could integrate , by making a substitution.

Anyway what would be the hyperbolic sub, for this involving coshu.

I don't have my textbook with me. I am at work, and I am trying to finish my homework.
 
Need i say that your initial approach is incorrect...?

It also works with [itex]\sec u[/itex]...

Daniel.
 
whozum said:
Why is it that everyone has been taught about hyperbolic trig functions when we skipped it in high school and calc 1-3?

I wouldn't know.Self-taught hyperbolic trigonometry in HS (XII-th grade) when i had to evaluate antiderivatives using substitution...

Daniel.
 
Ok, then,

Let x = secu

dx/du = secu.tanu

sec^2u-tan^2u = 1

dy/dx = 1/secu.tanu


Then what, to do!
 
[tex]\int \frac{\sec u \ \tan u \ du}{\sec u \ \tan u} =\int du = u+C = \mbox{arcsec} \ x +C[/tex]

Daniel.
 
Ok, I applied the initial conditions:

y = pi when x = 2

and I got,

y = sec-1(x) + pi - sec-1(2)

can we write this as:

y = sec-1 (x/2) + pi
 
  • #10
"sec" (as "cosine") is multivalued.But your could take

[tex]\sec^{-1} 2=\frac{\pi}{3}[/tex]

Daniel.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K