Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 6 x - 7\right)^{5} e^{- x} \cos^{8}{\left(x \right)}}{\left(3 x - 8\right)^{8} \sin^{5}{\left(x \right)}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(- 6 x - 7\right)^{5} e^{- x} \cos^{8}{\left(x \right)}}{\left(3 x - 8\right)^{8} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(- 6 x - 7 \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)}- x - 8 \ln{\left(3 x - 8 \right)} - 5 \ln{\left(\sin{\left(x \right)} \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{24}{3 x - 8} - \frac{30}{- 6 x - 7} 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{24}{3 x - 8} - \frac{30}{- 6 x - 7}\right)\left(\frac{\left(- 6 x - 7\right)^{5} e^{- x} \cos^{8}{\left(x \right)}}{\left(3 x - 8\right)^{8} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} - \frac{30}{- 6 x - 7}-1 - \frac{5}{\tan{\left(x \right)}} - \frac{24}{3 x - 8}\right)\left(\frac{\left(- 6 x - 7\right)^{5} e^{- x} \cos^{8}{\left(x \right)}}{\left(3 x - 8\right)^{8} \sin^{5}{\left(x \right)}} \right)