Answer
$$ y'
=\frac{x(x^2+1)}{\sqrt{x+1}}\left( \frac{1}{ x}+\frac{2x}{ x^2+1}-\frac{1}{2}\frac{1}{x+1}\right).$$
Work Step by Step
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
$$ y'=y\left( \frac{1}{ x}+\frac{2x}{ x^2+1}-\frac{1}{2}\frac{1}{x+1}\right)\\
=\frac{x(x^2+1)}{\sqrt{x+1}}\left( \frac{1}{ x}+\frac{2x}{ x^2+1}-\frac{1}{2}\frac{1}{x+1}\right).$$
