Find the derivative of $LaTeX: \displaystyle y = \frac{\left(2 x - 7\right)^{7} \left(3 x + 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(9 x + 9\right)^{4} \sqrt{\left(x + 7\right)^{7}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(2 x - 7\right)^{7} \left(3 x + 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(9 x + 9\right)^{4} \sqrt{\left(x + 7\right)^{7}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 7 \ln{\left(2 x - 7 \right)} + 3 \ln{\left(3 x + 5 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- \frac{7 \ln{\left(x + 7 \right)}}{2} - 4 \ln{\left(9 x + 9 \right)}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = 1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{36}{9 x + 9} + \frac{9}{3 x + 5} + \frac{14}{2 x - 7} - \frac{7}{2 \left(x + 7\right)}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{36}{9 x + 9} + \frac{9}{3 x + 5} + \frac{14}{2 x - 7} - \frac{7}{2 \left(x + 7\right)}\right)\left(\frac{\left(2 x - 7\right)^{7} \left(3 x + 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(9 x + 9\right)^{4} \sqrt{\left(x + 7\right)^{7}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(1 + \frac{6}{\tan{\left(x \right)}} + \frac{9}{3 x + 5} + \frac{14}{2 x - 7}- \frac{36}{9 x + 9} - \frac{7}{2 \left(x + 7\right)}\right)\left(\frac{\left(2 x - 7\right)^{7} \left(3 x + 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(9 x + 9\right)^{4} \sqrt{\left(x + 7\right)^{7}}} \right)$