Find the derivative of LaTeX: \displaystyle y = \frac{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- 6 x - 7\right)^{6} \left(2 x + 2\right)^{7} \sqrt{\left(7 x + 9\right)^{5}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- 6 x - 7\right)^{6} \left(2 x + 2\right)^{7} \sqrt{\left(7 x + 9\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(\sin{\left(x \right)} \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(- 6 x - 7 \right)} - 7 \ln{\left(2 x + 2 \right)} - \frac{5 \ln{\left(7 x + 9 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{2 \left(7 x + 9\right)} - \frac{14}{2 x + 2} + \frac{36}{- 6 x - 7} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{2 \left(7 x + 9\right)} - \frac{14}{2 x + 2} + \frac{36}{- 6 x - 7}\right)\left(\frac{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- 6 x - 7\right)^{6} \left(2 x + 2\right)^{7} \sqrt{\left(7 x + 9\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{2}{\tan{\left(x \right)}}- \frac{35}{2 \left(7 x + 9\right)} - \frac{14}{2 x + 2} + \frac{36}{- 6 x - 7}\right)\left(\frac{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- 6 x - 7\right)^{6} \left(2 x + 2\right)^{7} \sqrt{\left(7 x + 9\right)^{5}}} \right)