Simplifying an exponential with a square root

[Mr Davis 97]

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 $x \ge \frac{1}{2}$.

 

[Orodruin]

Was there something wrong with $\sqrt{|2x-1|}$?

 

[Mr Davis 97]

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 $\sqrt{|2x-1|}$

 

[JonnyG]

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 $y = 0$

 

[Orodruin]

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.