Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{\left(9 x + 1\right)^{7}} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 5 x - 8\right)^{3} \left(x + 9\right)^{4}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\sqrt{\left(9 x + 1\right)^{7}} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 5 x - 8\right)^{3} \left(x + 9\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(9 x + 1 \right)}}{2} + 2 \ln{\left(\sin{\left(x \right)} \right)}- x - 3 \ln{\left(- 5 x - 8 \right)} - 4 \ln{\left(x + 9 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{63}{2 \left(9 x + 1\right)} - \frac{4}{x + 9} + \frac{15}{- 5 x - 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{63}{2 \left(9 x + 1\right)} - \frac{4}{x + 9} + \frac{15}{- 5 x - 8}\right)\left(\frac{\sqrt{\left(9 x + 1\right)^{7}} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 5 x - 8\right)^{3} \left(x + 9\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{2}{\tan{\left(x \right)}} + \frac{63}{2 \left(9 x + 1\right)}-1 - \frac{4}{x + 9} + \frac{15}{- 5 x - 8}\right)\left(\frac{\sqrt{\left(9 x + 1\right)^{7}} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 5 x - 8\right)^{3} \left(x + 9\right)^{4}} \right)