Answer
$$y'=-\frac{3x+2}{2x(x+1)}$$
Work Step by Step
$$y=\ln\frac{1}{x\sqrt{x+1}}$$
We have $$y'=\frac{1}{\frac{1}{x\sqrt{x+1}}}\Big(\frac{1}{x\sqrt{x+1}}\Big)'$$
$$y'=x\sqrt{x+1}\times\frac{(1)'x\sqrt{x+1}-1(x\sqrt{x+1})'}{x^2(x+1)}$$
$$y'=x\sqrt{x+1}\times\frac{0\times x\sqrt{x+1}-\Big(\sqrt{x+1}+x(\sqrt{x+1})'\Big)}{x^2(x+1)}$$
$$y'=x\sqrt{x+1}\times\frac{-\Big(\sqrt{x+1}+x(\sqrt{x+1})'\Big)}{x^2(x+1)}$$
For $(\sqrt{x+1})'$: $(\sqrt{x+1})'=\Big((x+1)^{1/2}\Big)'=\frac{1}{2}(x+1)^{-1/2}(x+1)'=\frac{1}{2\sqrt{x+1}}$
$$y'=x\sqrt{x+1}\times\frac{-\Big(\sqrt{x+1}+\frac{x}{2\sqrt{x+1}}\Big)}{x^2(x+1)}$$
$$y'=\frac{-x\sqrt{x+1}\times\Big(\frac{2(x+1)+x}{2\sqrt{x+1}}\Big)}{x^2(x+1)}$$
$$y'=\frac{-\sqrt{x+1}\times\frac{3x+2}{2\sqrt{x+1}}}{x(x+1)}$$
$$y'=\frac{-\frac{3x+2}{2}}{x(x+1)}$$
$$y'=-\frac{3x+2}{2x(x+1)}$$
