Find the derivative of $LaTeX: \displaystyle y = \frac{\left(5 - 4 x\right)^{6} \left(6 x + 1\right)^{6} e^{x} \sin^{7}{\left(x \right)}}{\left(x - 4\right)^{7} \left(9 x - 7\right)^{5} \sqrt{\left(4 x + 7\right)^{7}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(5 - 4 x\right)^{6} \left(6 x + 1\right)^{6} e^{x} \sin^{7}{\left(x \right)}}{\left(x - 4\right)^{7} \left(9 x - 7\right)^{5} \sqrt{\left(4 x + 7\right)^{7}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 6 \ln{\left(5 - 4 x \right)} + 6 \ln{\left(6 x + 1 \right)} + 7 \ln{\left(\sin{\left(x \right)} \right)}- 7 \ln{\left(x - 4 \right)} - \frac{7 \ln{\left(4 x + 7 \right)}}{2} - 5 \ln{\left(9 x - 7 \right)}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{45}{9 x - 7} + \frac{36}{6 x + 1} - \frac{14}{4 x + 7} - \frac{7}{x - 4} - \frac{24}{5 - 4 x}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{45}{9 x - 7} + \frac{36}{6 x + 1} - \frac{14}{4 x + 7} - \frac{7}{x - 4} - \frac{24}{5 - 4 x}\right)\left(\frac{\left(5 - 4 x\right)^{6} \left(6 x + 1\right)^{6} e^{x} \sin^{7}{\left(x \right)}}{\left(x - 4\right)^{7} \left(9 x - 7\right)^{5} \sqrt{\left(4 x + 7\right)^{7}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(1 + \frac{7}{\tan{\left(x \right)}} + \frac{36}{6 x + 1} - \frac{24}{5 - 4 x}- \frac{45}{9 x - 7} - \frac{14}{4 x + 7} - \frac{7}{x - 4}\right)\left(\frac{\left(5 - 4 x\right)^{6} \left(6 x + 1\right)^{6} e^{x} \sin^{7}{\left(x \right)}}{\left(x - 4\right)^{7} \left(9 x - 7\right)^{5} \sqrt{\left(4 x + 7\right)^{7}}} \right)$