Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 - 6 x\right)^{4} \left(7 x - 9\right)^{8} e^{- x} \cos^{8}{\left(x \right)}}{\left(x - 7\right)^{3} \left(2 x + 3\right)^{6} \sin^{3}{\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(7 - 6 x\right)^{4} \left(7 x - 9\right)^{8} e^{- x} \cos^{8}{\left(x \right)}}{\left(x - 7\right)^{3} \left(2 x + 3\right)^{6} \sin^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(7 - 6 x \right)} + 8 \ln{\left(7 x - 9 \right)} + 8 \ln{\left(\cos{\left(x \right)} \right)}- x - 3 \ln{\left(x - 7 \right)} - 6 \ln{\left(2 x + 3 \right)} - 3 \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{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{56}{7 x - 9} - \frac{12}{2 x + 3} - \frac{3}{x - 7} - \frac{24}{7 - 6 x} 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{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{56}{7 x - 9} - \frac{12}{2 x + 3} - \frac{3}{x - 7} - \frac{24}{7 - 6 x}\right)\left(\frac{\left(7 - 6 x\right)^{4} \left(7 x - 9\right)^{8} e^{- x} \cos^{8}{\left(x \right)}}{\left(x - 7\right)^{3} \left(2 x + 3\right)^{6} \sin^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} + \frac{56}{7 x - 9} - \frac{24}{7 - 6 x}-1 - \frac{3}{\tan{\left(x \right)}} - \frac{12}{2 x + 3} - \frac{3}{x - 7}\right)\left(\frac{\left(7 - 6 x\right)^{4} \left(7 x - 9\right)^{8} e^{- x} \cos^{8}{\left(x \right)}}{\left(x - 7\right)^{3} \left(2 x + 3\right)^{6} \sin^{3}{\left(x \right)}} \right)