Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 9\right)^{4} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(6 x + 8\right)^{3} \sqrt{\left(3 x + 9\right)^{5}}}

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 + 9\right)^{4} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(6 x + 8\right)^{3} \sqrt{\left(3 x + 9\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(x + 9 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)}- x - \frac{5 \ln{\left(3 x + 9 \right)}}{2} - 3 \ln{\left(6 x + 8 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{6 x + 8} - \frac{15}{2 \left(3 x + 9\right)} + \frac{4}{x + 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{18}{6 x + 8} - \frac{15}{2 \left(3 x + 9\right)} + \frac{4}{x + 9}\right)\left(\frac{\left(x + 9\right)^{4} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(6 x + 8\right)^{3} \sqrt{\left(3 x + 9\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} + \frac{5}{\tan{\left(x \right)}} + \frac{4}{x + 9}-1 - \frac{18}{6 x + 8} - \frac{15}{2 \left(3 x + 9\right)}\right)\left(\frac{\left(x + 9\right)^{4} e^{- x} \sin^{5}{\left(x \right)} \cos^{8}{\left(x \right)}}{\left(6 x + 8\right)^{3} \sqrt{\left(3 x + 9\right)^{5}}} \right)