Question

Find the derivative of the given function

Answer 1

\[g(x) =\sqrt{\frac{ 2x+1 }{ 2x-1 }}\]

Answer 2

\[\Large\rm \color{royalblue}{g'(x)}=\frac{1}{2\sqrt{\frac{2x+1}{2x-1}}}\color{royalblue}{\left(\frac{2x+1}{2x-1}\right)'}\]So we take our square root derivative, and we then we have to chain.

Answer 3

Do you remember your sqrt(x) derivative?
\(\Large\rm \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt{x}}\)
Good one to memorize.

You can instead write it as 1/2 power, and apply power rule if you like.

Answer 4

Does this step make sense?\[\Large\rm \color{royalblue}{g'(x)}=\frac{1}{2}\sqrt{\frac{2x-1}{2x+1}}\color{royalblue}{\left(\frac{2x+1}{2x-1}\right)'}\]If I do this?

Answer 5

yes

Answer 6

So it looks like you'll need to apply quotient rule for the chain.
Here is how the setup will look:
\[\large\rm \color{royalblue}{g'(x)}=\frac{1}{2}\sqrt{\frac{2x-1}{2x+1}}\left(\frac{\color{royalblue}{(2x+1)'}(2x-1)+(2x+1)\color{royalblue}{(2x-1)'}}{(2x-1)^2}\right)\]

Answer 7

Mmmm what do you think?
Confusing?

Can you handle it from there? :D

Answer 8

yes thank you