Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 3\right)^{5} \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(6 x + 7\right)^{6} \sqrt{\left(3 x + 9\right)^{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 - 3\right)^{5} \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(6 x + 7\right)^{6} \sqrt{\left(3 x + 9\right)^{3}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(x - 3 \right)} + 3 \ln{\left(\sin{\left(x \right)} \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- \frac{3 \ln{\left(3 x + 9 \right)}}{2} - 6 \ln{\left(6 x + 7 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{36}{6 x + 7} - \frac{9}{2 \left(3 x + 9\right)} + \frac{5}{x - 3} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{36}{6 x + 7} - \frac{9}{2 \left(3 x + 9\right)} + \frac{5}{x - 3}\right)\left(\frac{\left(x - 3\right)^{5} \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(6 x + 7\right)^{6} \sqrt{\left(3 x + 9\right)^{3}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{3}{\tan{\left(x \right)}} + \frac{5}{x - 3}- \frac{36}{6 x + 7} - \frac{9}{2 \left(3 x + 9\right)}\right)\left(\frac{\left(x - 3\right)^{5} \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(6 x + 7\right)^{6} \sqrt{\left(3 x + 9\right)^{3}}} \right)