Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 8 x - 2\right)^{2} \left(x + 6\right)^{7} \sqrt{5 x + 2} e^{x}}{\left(x - 9\right)^{8} \sin^{7}{\left(x \right)} \cos^{7}{\left(x \right)}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(- 8 x - 2\right)^{2} \left(x + 6\right)^{7} \sqrt{5 x + 2} e^{x}}{\left(x - 9\right)^{8} \sin^{7}{\left(x \right)} \cos^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 2 \ln{\left(- 8 x - 2 \right)} + 7 \ln{\left(x + 6 \right)} + \frac{\ln{\left(5 x + 2 \right)}}{2}- 8 \ln{\left(x - 9 \right)} - 7 \ln{\left(\sin{\left(x \right)} \right)} - 7 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{5}{2 \left(5 x + 2\right)} + \frac{7}{x + 6} - \frac{8}{x - 9} - \frac{16}{- 8 x - 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{5}{2 \left(5 x + 2\right)} + \frac{7}{x + 6} - \frac{8}{x - 9} - \frac{16}{- 8 x - 2}\right)\left(\frac{\left(- 8 x - 2\right)^{2} \left(x + 6\right)^{7} \sqrt{5 x + 2} e^{x}}{\left(x - 9\right)^{8} \sin^{7}{\left(x \right)} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{5}{2 \left(5 x + 2\right)} + \frac{7}{x + 6} - \frac{16}{- 8 x - 2}7 \tan{\left(x \right)} - \frac{7}{\tan{\left(x \right)}} - \frac{8}{x - 9}\right)\left(\frac{\left(- 8 x - 2\right)^{2} \left(x + 6\right)^{7} \sqrt{5 x + 2} e^{x}}{\left(x - 9\right)^{8} \sin^{7}{\left(x \right)} \cos^{7}{\left(x \right)}} \right)