Find the derivative of LaTeX: \displaystyle y = \frac{\left(8 - 3 x\right)^{7} \sqrt{\left(8 x + 9\right)^{5}} e^{- x} \sin^{4}{\left(x \right)}}{343 x^{3} \left(2 x - 7\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{\left(8 - 3 x\right)^{7} \sqrt{\left(8 x + 9\right)^{5}} e^{- x} \sin^{4}{\left(x \right)}}{343 x^{3} \left(2 x - 7\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(8 - 3 x \right)} + \frac{5 \ln{\left(8 x + 9 \right)}}{2} + 4 \ln{\left(\sin{\left(x \right)} \right)}- x - 3 \ln{\left(x \right)} - 4 \ln{\left(2 x - 7 \right)} - 3 \ln{\left(7 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{8 x + 9} - \frac{8}{2 x - 7} - \frac{21}{8 - 3 x} - \frac{3}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{8 x + 9} - \frac{8}{2 x - 7} - \frac{21}{8 - 3 x} - \frac{3}{x}\right)\left(\frac{\left(8 - 3 x\right)^{7} \sqrt{\left(8 x + 9\right)^{5}} e^{- x} \sin^{4}{\left(x \right)}}{343 x^{3} \left(2 x - 7\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{4}{\tan{\left(x \right)}} + \frac{20}{8 x + 9} - \frac{21}{8 - 3 x}-1 - \frac{8}{2 x - 7} - \frac{3}{x}\right)\left(\frac{\left(8 - 3 x\right)^{7} \sqrt{\left(8 x + 9\right)^{5}} e^{- x} \sin^{4}{\left(x \right)}}{343 x^{3} \left(2 x - 7\right)^{4}} \right)