Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 5 x - 9\right)^{2} \left(- x - 4\right)^{7} \left(x + 9\right)^{2} e^{x}}{49 x^{2} \sin^{4}{\left(x \right)} \cos^{6}{\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(- 5 x - 9\right)^{2} \left(- x - 4\right)^{7} \left(x + 9\right)^{2} e^{x}}{49 x^{2} \sin^{4}{\left(x \right)} \cos^{6}{\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(- 5 x - 9 \right)} + 7 \ln{\left(- x - 4 \right)} + 2 \ln{\left(x + 9 \right)}- 2 \ln{\left(x \right)} - 4 \ln{\left(\sin{\left(x \right)} \right)} - 6 \ln{\left(\cos{\left(x \right)} \right)} - 2 \ln{\left(7 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{2}{x + 9} - \frac{7}{- x - 4} - \frac{10}{- 5 x - 9} - \frac{2}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{2}{x + 9} - \frac{7}{- x - 4} - \frac{10}{- 5 x - 9} - \frac{2}{x}\right)\left(\frac{\left(- 5 x - 9\right)^{2} \left(- x - 4\right)^{7} \left(x + 9\right)^{2} e^{x}}{49 x^{2} \sin^{4}{\left(x \right)} \cos^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{2}{x + 9} - \frac{7}{- x - 4} - \frac{10}{- 5 x - 9}6 \tan{\left(x \right)} - \frac{4}{\tan{\left(x \right)}} - \frac{2}{x}\right)\left(\frac{\left(- 5 x - 9\right)^{2} \left(- x - 4\right)^{7} \left(x + 9\right)^{2} e^{x}}{49 x^{2} \sin^{4}{\left(x \right)} \cos^{6}{\left(x \right)}} \right)