Use a stated interval to choose the square-root sign

The chain rule gives $-\cos x/\sqrt{1-\sin^2x}=-\cos x/\sqrt{\cos^2x}$. On the stated interval $\cos x<0$, so $\sqrt{\cos^2x}=|\cos x|=-\cos x$. The derivative is therefore $1$.

On the real domain $x>0$, rewrite as $\frac12[\ln(e^x+1)-\ln(1-e^{-x})]$. Differentiate to get $\frac12\left[\dfrac{e^x}{e^x+1}-\dfrac{e^{-x}}{1-e^{-x}}\right]=\frac12\left[\dfrac{e^x}{e^x+1}-\dfrac1{e^x-1}\right]$.