Answer
$$y'=\frac{1}{2}\sqrt{\frac{x(x+2)}{(2x+1)(3x+2)}}\left(\frac{1}{ x}+ \frac{1}{ x+2}-\frac{2}{ 2x+1}-\frac{3}{3x+2}\right)$$
Work Step by Step
Taking the $\ln $ of both sides of the equation, we get
$$\ln y= \ln\sqrt{\frac{x(x+2)}{(2x+1)(3x+2)}}$$
Then using the properties of $\ln $, we have
$$\ln y=\frac{1}{2} \ln \frac{x(x+2)}{(2x+1)(3x+2)}\\
=\frac{1}{2}(\ln x+\ln (x+2)-\ln(2x+1)-\ln (3x+2)).$$
Now taking the derivative for the above equation, we have
$$\frac{y'}{y}=\frac{1}{2} (\frac{1}{ x}+ \frac{1}{ x+2}-\frac{2}{ 2x+1}-\frac{3}{3x+2}),$$
Hence $ y'$ is given by
$$ y'=\frac{y}{2}\left(\frac{1}{ x}+ \frac{1}{ x+2}-\frac{2}{ 2x+1}-\frac{3}{3x+2}\right).\\
y'=\frac{1}{2}\sqrt{\frac{x(x+2)}{(2x+1)(3x+2)}}\left(\frac{1}{ x}+ \frac{1}{ x+2}-\frac{2}{ 2x+1}-\frac{3}{3x+2}\right)$$
