Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 x - 8\right)^{7} e^{- x} \cos^{6}{\left(x \right)}}{\left(2 x - 6\right)^{7} \left(2 x + 8\right)^{2} \sqrt{\left(8 x + 6\right)^{3}}}

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 - 8\right)^{7} e^{- x} \cos^{6}{\left(x \right)}}{\left(2 x - 6\right)^{7} \left(2 x + 8\right)^{2} \sqrt{\left(8 x + 6\right)^{3}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(6 x - 8 \right)} + 6 \ln{\left(\cos{\left(x \right)} \right)}- x - 7 \ln{\left(2 x - 6 \right)} - 2 \ln{\left(2 x + 8 \right)} - \frac{3 \ln{\left(8 x + 6 \right)}}{2} 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 + 6} + \frac{42}{6 x - 8} - \frac{4}{2 x + 8} - \frac{14}{2 x - 6} 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 + 6} + \frac{42}{6 x - 8} - \frac{4}{2 x + 8} - \frac{14}{2 x - 6}\right)\left(\frac{\left(6 x - 8\right)^{7} e^{- x} \cos^{6}{\left(x \right)}}{\left(2 x - 6\right)^{7} \left(2 x + 8\right)^{2} \sqrt{\left(8 x + 6\right)^{3}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 6 \tan{\left(x \right)} + \frac{42}{6 x - 8}-1 - \frac{12}{8 x + 6} - \frac{4}{2 x + 8} - \frac{14}{2 x - 6}\right)\left(\frac{\left(6 x - 8\right)^{7} e^{- x} \cos^{6}{\left(x \right)}}{\left(2 x - 6\right)^{7} \left(2 x + 8\right)^{2} \sqrt{\left(8 x + 6\right)^{3}}} \right)