Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 6 x - 3\right)^{7} \left(x + 1\right)^{7} e^{x}}{\left(- 5 x - 6\right)^{5} \sqrt{\left(4 x + 2\right)^{7}} \cos^{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(- 6 x - 3\right)^{7} \left(x + 1\right)^{7} e^{x}}{\left(- 5 x - 6\right)^{5} \sqrt{\left(4 x + 2\right)^{7}} \cos^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 7 \ln{\left(- 6 x - 3 \right)} + 7 \ln{\left(x + 1 \right)}- 5 \ln{\left(- 5 x - 6 \right)} - \frac{7 \ln{\left(4 x + 2 \right)}}{2} - 3 \ln{\left(\cos{\left(x \right)} \right)} 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{14}{4 x + 2} + \frac{7}{x + 1} + \frac{25}{- 5 x - 6} - \frac{42}{- 6 x - 3} 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{14}{4 x + 2} + \frac{7}{x + 1} + \frac{25}{- 5 x - 6} - \frac{42}{- 6 x - 3}\right)\left(\frac{\left(- 6 x - 3\right)^{7} \left(x + 1\right)^{7} e^{x}}{\left(- 5 x - 6\right)^{5} \sqrt{\left(4 x + 2\right)^{7}} \cos^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{7}{x + 1} - \frac{42}{- 6 x - 3}3 \tan{\left(x \right)} - \frac{14}{4 x + 2} + \frac{25}{- 5 x - 6}\right)\left(\frac{\left(- 6 x - 3\right)^{7} \left(x + 1\right)^{7} e^{x}}{\left(- 5 x - 6\right)^{5} \sqrt{\left(4 x + 2\right)^{7}} \cos^{3}{\left(x \right)}} \right)