Find the derivative of $LaTeX: \displaystyle y = \frac{\left(- 9 x - 1\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{\left(7 - 8 x\right)^{2} \left(x + 7\right)^{2} \sqrt{\left(4 x + 7\right)^{3}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(- 9 x - 1\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{\left(7 - 8 x\right)^{2} \left(x + 7\right)^{2} \sqrt{\left(4 x + 7\right)^{3}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 3 \ln{\left(- 9 x - 1 \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(7 - 8 x \right)} - 2 \ln{\left(x + 7 \right)} - \frac{3 \ln{\left(4 x + 7 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{6}{4 x + 7} - \frac{2}{x + 7} - \frac{27}{- 9 x - 1} + \frac{16}{7 - 8 x}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{6}{4 x + 7} - \frac{2}{x + 7} - \frac{27}{- 9 x - 1} + \frac{16}{7 - 8 x}\right)\left(\frac{\left(- 9 x - 1\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{\left(7 - 8 x\right)^{2} \left(x + 7\right)^{2} \sqrt{\left(4 x + 7\right)^{3}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(- 4 \tan{\left(x \right)} + 1 - \frac{27}{- 9 x - 1}- \frac{6}{4 x + 7} - \frac{2}{x + 7} + \frac{16}{7 - 8 x}\right)\left(\frac{\left(- 9 x - 1\right)^{3} e^{x} \cos^{4}{\left(x \right)}}{\left(7 - 8 x\right)^{2} \left(x + 7\right)^{2} \sqrt{\left(4 x + 7\right)^{3}}} \right)$