Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{x + 5} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{2} \left(6 x + 2\right)^{7}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\sqrt{x + 5} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{2} \left(6 x + 2\right)^{7}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{\ln{\left(x + 5 \right)}}{2} + 2 \ln{\left(\sin{\left(x \right)} \right)}- 2 \ln{\left(x + 3 \right)} - 7 \ln{\left(6 x + 2 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{42}{6 x + 2} + \frac{1}{2 \left(x + 5\right)} - \frac{2}{x + 3} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{42}{6 x + 2} + \frac{1}{2 \left(x + 5\right)} - \frac{2}{x + 3}\right)\left(\frac{\sqrt{x + 5} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{2} \left(6 x + 2\right)^{7}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{2}{\tan{\left(x \right)}} + \frac{1}{2 \left(x + 5\right)}- \frac{42}{6 x + 2} - \frac{2}{x + 3}\right)\left(\frac{\sqrt{x + 5} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{2} \left(6 x + 2\right)^{7}} \right)