Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 4 x - 6\right)^{3} \sqrt{\left(8 x + 9\right)^{3}} e^{- x} \cos^{6}{\left(x \right)}}{\left(4 - 4 x\right)^{6} \left(2 x - 9\right)^{6}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(- 4 x - 6\right)^{3} \sqrt{\left(8 x + 9\right)^{3}} e^{- x} \cos^{6}{\left(x \right)}}{\left(4 - 4 x\right)^{6} \left(2 x - 9\right)^{6}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(- 4 x - 6 \right)} + \frac{3 \ln{\left(8 x + 9 \right)}}{2} + 6 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(4 - 4 x \right)} - 6 \ln{\left(2 x - 9 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{12}{8 x + 9} - \frac{12}{2 x - 9} - \frac{12}{- 4 x - 6} + \frac{24}{4 - 4 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{12}{8 x + 9} - \frac{12}{2 x - 9} - \frac{12}{- 4 x - 6} + \frac{24}{4 - 4 x}\right)\left(\frac{\left(- 4 x - 6\right)^{3} \sqrt{\left(8 x + 9\right)^{3}} e^{- x} \cos^{6}{\left(x \right)}}{\left(4 - 4 x\right)^{6} \left(2 x - 9\right)^{6}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 6 \tan{\left(x \right)} + \frac{12}{8 x + 9} - \frac{12}{- 4 x - 6}-1 - \frac{12}{2 x - 9} + \frac{24}{4 - 4 x}\right)\left(\frac{\left(- 4 x - 6\right)^{3} \sqrt{\left(8 x + 9\right)^{3}} e^{- x} \cos^{6}{\left(x \right)}}{\left(4 - 4 x\right)^{6} \left(2 x - 9\right)^{6}} \right)