Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 - 2 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 2 x - 5\right)^{5} \sqrt{\left(9 x + 6\right)^{3}} \cos^{7}{\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 - 2 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 2 x - 5\right)^{5} \sqrt{\left(9 x + 6\right)^{3}} \cos^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(6 - 2 x \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- x - 5 \ln{\left(- 2 x - 5 \right)} - \frac{3 \ln{\left(9 x + 6 \right)}}{2} - 7 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{27}{2 \left(9 x + 6\right)} + \frac{10}{- 2 x - 5} - \frac{8}{6 - 2 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{27}{2 \left(9 x + 6\right)} + \frac{10}{- 2 x - 5} - \frac{8}{6 - 2 x}\right)\left(\frac{\left(6 - 2 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 2 x - 5\right)^{5} \sqrt{\left(9 x + 6\right)^{3}} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{2}{\tan{\left(x \right)}} - \frac{8}{6 - 2 x}7 \tan{\left(x \right)} - 1 - \frac{27}{2 \left(9 x + 6\right)} + \frac{10}{- 2 x - 5}\right)\left(\frac{\left(6 - 2 x\right)^{4} e^{- x} \sin^{2}{\left(x \right)}}{\left(- 2 x - 5\right)^{5} \sqrt{\left(9 x + 6\right)^{3}} \cos^{7}{\left(x \right)}} \right)