Differentiate a square root of inverse cosine and a logarithmic radical

Apply the square-root chain rule and then inverse cosine: $\frac12[\cos^{-1}(x/2)]^{-1/2}\dfrac{-1/2}{\sqrt{1-(x/2)^2}}=-\dfrac1{2\sqrt{(4-x^2)\cos^{-1}(x/2)}}$. The derivative is defined for $-2<x<2$.

The real logarithm requires $x>1$. Then $\sqrt{(x+1)^3/(x^2-1)}=(x+1)/\sqrt{x-1}$, since $\sqrt{(x+1)^2}=x+1$ on this domain. Differentiate $\ln(x+1)-\frac12\ln(x-1)$ to obtain $\dfrac1{x+1}-\dfrac1{2(x-1)}=\dfrac{x-3}{2(x^2-1)}$.

The source labels this longer route not recommended. Let $u=(x+1)^3/(x^2-1)$. Then $u\prime=\dfrac{3(x+1)^2(x^2-1)-2x(x+1)^3}{(x^2-1)^2}$. Applying the log and square-root chain rules gives $u\prime/(2u)=\dfrac{x-3}{2(x^2-1)}$.