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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(9 - 4 x\right)^{4} \left(4 x + 6\right)^{7} e^{x} \sin^{3}{\left(x \right)}}{\left(x - 9\right)^{2} \left(9 x - 9\right)^{7} \sqrt{\left(4 x + 4\right)^{5}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 4 \ln{\left(9 - 4 x \right)} + 7 \ln{\left(4 x + 6 \right)} + 3 \ln{\left(\sin{\left(x \right)} \right)}- 2 \ln{\left(x - 9 \right)} - \frac{5 \ln{\left(4 x + 4 \right)}}{2} - 7 \ln{\left(9 x - 9 \right)}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = 1 + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{63}{9 x - 9} + \frac{28}{4 x + 6} - \frac{10}{4 x + 4} - \frac{2}{x - 9} - \frac{16}{9 - 4 x}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(1 + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{63}{9 x - 9} + \frac{28}{4 x + 6} - \frac{10}{4 x + 4} - \frac{2}{x - 9} - \frac{16}{9 - 4 x}\right)\left(\frac{\left(9 - 4 x\right)^{4} \left(4 x + 6\right)^{7} e^{x} \sin^{3}{\left(x \right)}}{\left(x - 9\right)^{2} \left(9 x - 9\right)^{7} \sqrt{\left(4 x + 4\right)^{5}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(1 + \frac{3}{\tan{\left(x \right)}} + \frac{28}{4 x + 6} - \frac{16}{9 - 4 x}- \frac{63}{9 x - 9} - \frac{10}{4 x + 4} - \frac{2}{x - 9}\right)\left(\frac{\left(9 - 4 x\right)^{4} \left(4 x + 6\right)^{7} e^{x} \sin^{3}{\left(x \right)}}{\left(x - 9\right)^{2} \left(9 x - 9\right)^{7} \sqrt{\left(4 x + 4\right)^{5}}} \right)$