Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 4\right)^{3} \sin^{5}{\left(x \right)}}{\left(x + 9\right)^{4} \left(5 x + 2\right)^{5} \sqrt{9 x + 3}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(x + 4\right)^{3} \sin^{5}{\left(x \right)}}{\left(x + 9\right)^{4} \left(5 x + 2\right)^{5} \sqrt{9 x + 3}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(x + 4 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(x + 9 \right)} - 5 \ln{\left(5 x + 2 \right)} - \frac{\ln{\left(9 x + 3 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{2 \left(9 x + 3\right)} - \frac{25}{5 x + 2} - \frac{4}{x + 9} + \frac{3}{x + 4} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{2 \left(9 x + 3\right)} - \frac{25}{5 x + 2} - \frac{4}{x + 9} + \frac{3}{x + 4}\right)\left(\frac{\left(x + 4\right)^{3} \sin^{5}{\left(x \right)}}{\left(x + 9\right)^{4} \left(5 x + 2\right)^{5} \sqrt{9 x + 3}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{5}{\tan{\left(x \right)}} + \frac{3}{x + 4}- \frac{9}{2 \left(9 x + 3\right)} - \frac{25}{5 x + 2} - \frac{4}{x + 9}\right)\left(\frac{\left(x + 4\right)^{3} \sin^{5}{\left(x \right)}}{\left(x + 9\right)^{4} \left(5 x + 2\right)^{5} \sqrt{9 x + 3}} \right)