Answer
$$ y'= (2x+1)(4x^2)\sqrt{x-9})\left( \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9}\right).$$
Work Step by Step
Taking the $\ln $ of both sides of the equation, we get
$$\ln y= \ln((2x+1)(4x^2)\sqrt{x-9})$$ Then using the properties of $\ln $, we have $$\ln y= \ln (2x+1)+\ln (2x)^2+\frac{1}{2}\ln(x-9)\\ =\ln (2x+1)+2\ln (2x)+\frac{1}{2}\ln(x-9).$$ Now taking the derivative for the above equation, we have
$$\frac{y'}{y}= \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9},$$ Hence $ y'$ is given by
$$ y'=y\left( \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9}\right)\\ =(2x+1)(4x^2)\sqrt{x-9})\left( \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9}\right).$$
