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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(7 - 4 x\right)^{2} e^{x} \cos^{4}{\left(x \right)}}{\left(- 5 x - 7\right)^{5} \sqrt{\left(9 x + 5\right)^{3}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = x + 2 \ln{\left(7 - 4 x \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- 5 \ln{\left(- 5 x - 7 \right)} - \frac{3 \ln{\left(9 x + 5 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{27}{2 \left(9 x + 5\right)} + \frac{25}{- 5 x - 7} - \frac{8}{7 - 4 x}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{27}{2 \left(9 x + 5\right)} + \frac{25}{- 5 x - 7} - \frac{8}{7 - 4 x}\right)\left(\frac{\left(7 - 4 x\right)^{2} e^{x} \cos^{4}{\left(x \right)}}{\left(- 5 x - 7\right)^{5} \sqrt{\left(9 x + 5\right)^{3}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(- 4 \tan{\left(x \right)} + 1 - \frac{8}{7 - 4 x}- \frac{27}{2 \left(9 x + 5\right)} + \frac{25}{- 5 x - 7}\right)\left(\frac{\left(7 - 4 x\right)^{2} e^{x} \cos^{4}{\left(x \right)}}{\left(- 5 x - 7\right)^{5} \sqrt{\left(9 x + 5\right)^{3}}} \right)$