Determine dy/dx of the following and simplify if possible

[DevonZA]

y=$sin^{-1}(\frac{x-1}{x+1})$

My attempt:

$\frac{dy}{dx}$ = $\frac{1}{\sqrt{1-(\frac{(x-1)}{(x+1)})^2}}$ . $\frac{d}{dx}(\frac{x-1}{x+1})$

= $\frac{1}{\sqrt{(\frac{(x+1)^2-(x-1)^2}{(x+1)^2}}}$ . $\frac{(x+1)(1)-(x-1)(1)}{(x+1)^2}$

= $\frac{x+1}{\sqrt{x^2+2x+1-x^2+2x-1}}$ . $\frac{(x+1-x+1)}{(x+1)^2}$

= $\frac{x+1-x+1}{(x+1)(\sqrt4x)}$

= $\frac{2}{(x+1)(\sqrt4x)}$

 

[PeroK]

It's correct except for a typo in your latex. It should be $\sqrt{4x}$

It's better to post in the homework section.

 

[DevonZA]

Final answer attached. Thanks to all who helped.