# B Simplifying an exponential with a square root

1. Jan 24, 2017

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

2. Jan 24, 2017

### Orodruin

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

3. Jan 24, 2017

### Mr Davis 97

Well, I then need to take the derivative of the resulting expression, and I don't see how to take the derivative of $\sqrt{|2x-1|}$

4. Jan 24, 2017

### 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 $y = 0$

5. Jan 24, 2017

### Orodruin

Staff Emeritus
Regardless of how you do things, your function will not be differentiable in x=1/2.