Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 9\right)^{2} e^{- x} \sin^{8}{\left(x \right)}}{\left(- 9 x - 2\right)^{4} \left(5 x - 3\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 - 9\right)^{2} e^{- x} \sin^{8}{\left(x \right)}}{\left(- 9 x - 2\right)^{4} \left(5 x - 3\right)^{3}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(x - 9 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)}- x - 4 \ln{\left(- 9 x - 2 \right)} - 3 \ln{\left(5 x - 3 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{5 x - 3} + \frac{2}{x - 9} + \frac{36}{- 9 x - 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{5 x - 3} + \frac{2}{x - 9} + \frac{36}{- 9 x - 2}\right)\left(\frac{\left(x - 9\right)^{2} e^{- x} \sin^{8}{\left(x \right)}}{\left(- 9 x - 2\right)^{4} \left(5 x - 3\right)^{3}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{8}{\tan{\left(x \right)}} + \frac{2}{x - 9}-1 - \frac{15}{5 x - 3} + \frac{36}{- 9 x - 2}\right)\left(\frac{\left(x - 9\right)^{2} e^{- x} \sin^{8}{\left(x \right)}}{\left(- 9 x - 2\right)^{4} \left(5 x - 3\right)^{3}} \right)