How Do I Apply L'Hopital's Rule to Exponential Derivatives?

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Homework Help Overview

The discussion revolves around applying L'Hopital's Rule to the limit of the expression \((2x - 1)^{1/(x-1)}\) as \(x\) approaches 1, particularly focusing on the derivative and the exponential form of the function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of L'Hopital's Rule and discuss the use of logarithms to simplify the limit. There are questions regarding the correct interpretation of the limit and the derivative rules applicable to the problem.

Discussion Status

The discussion is ongoing, with participants providing different interpretations of the limit's outcome. Some guidance has been offered regarding the use of logarithms and L'Hopital's Rule, but there is no consensus on the final result.

Contextual Notes

Participants note confusion regarding the application of L'Hopital's Rule and the correct limit evaluation, with references to textbook examples for clarification. There is mention of potential errors in understanding the problem setup.

KristinaMr
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Homework Statement
how can I find the derivative of (2x-1)^1/(x-1)...the second part is all in the exponent
Relevant Equations
I really don't know which rule applies in this case..maybe there's a way to rearrange the expression?..any help is appreciated
I encountered this problem is Hopital section...how do I even apply it?
 
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<br /> (2x - 1)^{1/(x-1)} = \exp\left(\frac{\ln (2x-1)}{x - 1}\right)<br />
 
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Do you mean L'Hopital? I don't know how that applies to this derivative either. Regarding the derivative, are you familiar with the Exponent rule for derivatives? (see Exponent Rule for Derivatives )
 
FactChecker said:
Do you mean L'Hopital? I don't know how that applies to this derivative either. Regarding the derivative, are you familiar with the Exponent rule for derivatives? (see Exponent Rule for Derivatives )
thank you for the exponent rule ..I somehow missed it

yea it was a limit of x-> 1 so that the exponent would be 1/0 ..the exercise said to apply L Hopital rule to solve ( by the way the result should be e)
 
KristinaMr said:
Problem Statement: how can I find the derivative of (2x-1)^1/(x-1)...the second part is all in the exponent
Relevant Equations: I really don't know which rule applies in this case..maybe there's a way to rearrange the expression?..any help is appreciated

I encountered this problem is Hopital section...how do I even apply it?
This problem has nothing to do with finding the derivative of that function.

Write the function above as an equation: ##y = (2x - 1)^{1/(x - 1)}##
Take the log (ln) of both sides to get ##\ln y = \frac 1 {x - 1} \ln(2x - 1) = \frac{\ln(2x - 1)}{x - 1}##
Now take the limit as x --> 1 of both sides. At this point the limit on the right side can be evaluated using L'Hopital's Rule.

Your textbook should have and example of this technique. Look at the example, and follow the steps in that example.

KristinaMr said:
by the way the result should be e
I don't think so, not as you have shown the problem. I get a value of ##e^2## for the limit.
 
Mark44 said:
I don't think so, not as you have shown the problem. I get a value of ##e^2## for the limit.
No, the limit is as given in the book, e1.
 
ehild said:
No, the limit is as given in the book, e1.
My mistake -- I can't read my own writing...
 

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