Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 7\right)^{4} \left(6 x - 7\right)^{6} e^{- x} \cos^{3}{\left(x \right)}}{\left(- 2 x - 1\right)^{7} \left(- x - 7\right)^{5} \sqrt{\left(2 x + 1\right)^{7}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(x - 7\right)^{4} \left(6 x - 7\right)^{6} e^{- x} \cos^{3}{\left(x \right)}}{\left(- 2 x - 1\right)^{7} \left(- x - 7\right)^{5} \sqrt{\left(2 x + 1\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(x - 7 \right)} + 6 \ln{\left(6 x - 7 \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)}- x - 7 \ln{\left(- 2 x - 1 \right)} - 5 \ln{\left(- x - 7 \right)} - \frac{7 \ln{\left(2 x + 1 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{36}{6 x - 7} - \frac{7}{2 x + 1} + \frac{4}{x - 7} + \frac{5}{- x - 7} + \frac{14}{- 2 x - 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{36}{6 x - 7} - \frac{7}{2 x + 1} + \frac{4}{x - 7} + \frac{5}{- x - 7} + \frac{14}{- 2 x - 1}\right)\left(\frac{\left(x - 7\right)^{4} \left(6 x - 7\right)^{6} e^{- x} \cos^{3}{\left(x \right)}}{\left(- 2 x - 1\right)^{7} \left(- x - 7\right)^{5} \sqrt{\left(2 x + 1\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{36}{6 x - 7} + \frac{4}{x - 7}-1 - \frac{7}{2 x + 1} + \frac{5}{- x - 7} + \frac{14}{- 2 x - 1}\right)\left(\frac{\left(x - 7\right)^{4} \left(6 x - 7\right)^{6} e^{- x} \cos^{3}{\left(x \right)}}{\left(- 2 x - 1\right)^{7} \left(- x - 7\right)^{5} \sqrt{\left(2 x + 1\right)^{7}}} \right)