Find the derivative of $LaTeX: \displaystyle y = \frac{\left(x + 4\right)^{2} \left(6 x + 8\right)^{3} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 7\right)^{5} \sqrt{\left(x + 3\right)^{5}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(\frac{\left(x + 4\right)^{2} \left(6 x + 8\right)^{3} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 7\right)^{5} \sqrt{\left(x + 3\right)^{5}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = 2 \ln{\left(x + 4 \right)} + 3 \ln{\left(6 x + 8 \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 5 \ln{\left(x - 7 \right)} - \frac{5 \ln{\left(x + 3 \right)}}{2}$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{18}{6 x + 8} + \frac{2}{x + 4} - \frac{5}{2 \left(x + 3\right)} - \frac{5}{x - 7}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{18}{6 x + 8} + \frac{2}{x + 4} - \frac{5}{2 \left(x + 3\right)} - \frac{5}{x - 7}\right)\left(\frac{\left(x + 4\right)^{2} \left(6 x + 8\right)^{3} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 7\right)^{5} \sqrt{\left(x + 3\right)^{5}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{18}{6 x + 8} + \frac{2}{x + 4}-1 - \frac{5}{2 \left(x + 3\right)} - \frac{5}{x - 7}\right)\left(\frac{\left(x + 4\right)^{2} \left(6 x + 8\right)^{3} e^{- x} \cos^{2}{\left(x \right)}}{\left(x - 7\right)^{5} \sqrt{\left(x + 3\right)^{5}}} \right)$