Simplifying an exponential with a square root

Click For Summary

Discussion Overview

The discussion revolves around the expression ##e^{\frac{1}{2} \log|2x-1|## and its simplification to ##\sqrt{2x-1}##. Participants explore the implications of this simplification, particularly concerning the domains of the functions involved and the differentiation of the resulting expression.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests that ##e^{\frac{1}{2} \log|2x-1|}## simplifies to ##\sqrt{2x-1}## but expresses uncertainty about justifying this due to differing domains.
  • Another participant questions whether there is an issue with using ##\sqrt{|2x-1|}## instead.
  • A later reply indicates a need to differentiate ##\sqrt{|2x-1|}## and suggests splitting into cases or using the chain rule with ##y = 2x - 1##, noting that ##|y|## is not differentiable at ##y = 0##.
  • One participant reiterates the challenge of differentiating ##\sqrt{|2x-1|}## and emphasizes that the function will not be differentiable at ##x = \frac{1}{2}## regardless of the approach taken.

Areas of Agreement / Disagreement

Participants express differing views on the simplification and differentiation of the expression, indicating that the discussion remains unresolved regarding the implications of the domains and differentiability.

Contextual Notes

Participants highlight limitations related to the domains of the functions and the differentiability at specific points, but do not resolve these issues.

Mr Davis 97
Messages
1,461
Reaction score
44
I have the expression ##e^{\frac{1}{2} \log|2x-1|}##. I am tempted to just say that this is equal to ##\sqrt{2x-1}## and be done with it. However, I am not sure how to justify this, since it seems that then the domains of the two functions would be different, since the latter would be all real numbers while the former would be ##x \ge \frac{1}{2}##.
 
Physics news on Phys.org
Was there something wrong with ##\sqrt{|2x-1|}##?
 
Orodruin said:
Was there something wrong with ##\sqrt{|2x-1|}##?
Well, I then need to take the derivative of the resulting expression, and I don't see how to take the derivative of ##\sqrt{|2x-1|}##
 
Split it into cases. Or set ##y = 2x -1## and differentiate ##\sqrt{\lvert y \rvert}## using the chain rule, remembering that ##\lvert y \rvert## is not differentiable when ##y = 0##
 
Mr Davis 97 said:
Well, I then need to take the derivative of the resulting expression, and I don't see how to take the derivative of ##\sqrt{|2x-1|}##
Regardless of how you do things, your function will not be differentiable in x=1/2.
 

Similar threads

High School Derivative of Square Root of x at 0
  • · Replies 3 ·
Replies
3
Views
6K
High School Is the sign of the square root dependent on the argument inside it?
  • · Replies 1 ·
Replies
1
Views
2K
High School Straightforward integration…
  • · Replies 27 ·
Replies
27
Views
3K
MHB Differentiation with square roots
  • · Replies 3 ·
Replies
3
Views
2K
High School Trigonometric substitution, a case I'd like to share
  • · Replies 3 ·
Replies
3
Views
4K
Undergrad Question about Inverse Derivative Hyperbola function
  • · Replies 2 ·
Replies
2
Views
1K
High School How do I invert this exponential function?
  • · Replies 3 ·
Replies
3
Views
4K
MHB How have I dropped a factor 2 on the square root of 19?
  • · Replies 1 ·
Replies
1
Views
2K
High School Question about this Lesson on Square Roots
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Different Substitution for Solving an Integral
  • · Replies 6 ·
Replies
6
Views
3K