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

[KristinaMr]

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?

 

[pasmith]

$$(2x - 1)^{1/(x-1)} = \exp\left(\frac{\ln (2x-1)}{x - 1}\right)$$

 

[FactChecker]

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 )

 

[KristinaMr]

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)

 

[Mark44]

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.

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.

 

[ehild]

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.

 

[Mark44]

No, the limit is as given in the book, e¹.

My mistake -- I can't read my own writing...