Calculus (3rd Edition)

$$y' =\frac{x(x^2+1)}{\sqrt{x+1}}\left( \frac{1}{ x}+\frac{2x}{ x^2+1}-\frac{1}{2}\frac{1}{x+1}\right).$$
Taking the $\ln$ of both sides of the equation, we get $$\ln y= \ln \frac{x(x^2+1)}{\sqrt{x+1}}$$ Then using the properties of $\ln$, we have $$\ln y= \ln x+\ln (x^2+1)-\frac{1}{2}\ln(x+1).$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= \frac{1}{ x}+\frac{2x}{ x^2+1}-\frac{1}{2}\frac{1}{x+1},$$ Hence $y'$ is given by $$y'=y\left( \frac{1}{ x}+\frac{2x}{ x^2+1}-\frac{1}{2}\frac{1}{x+1}\right)\\ =\frac{x(x^2+1)}{\sqrt{x+1}}\left( \frac{1}{ x}+\frac{2x}{ x^2+1}-\frac{1}{2}\frac{1}{x+1}\right).$$