Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 9\right)^{7} \left(4 x + 4\right)^{6} e^{x}}{\left(3 x - 5\right)^{7} \left(4 x + 9\right)^{5} \cos^{2}{\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(x - 9\right)^{7} \left(4 x + 4\right)^{6} e^{x}}{\left(3 x - 5\right)^{7} \left(4 x + 9\right)^{5} \cos^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 7 \ln{\left(x - 9 \right)} + 6 \ln{\left(4 x + 4 \right)}- 7 \ln{\left(3 x - 5 \right)} - 5 \ln{\left(4 x + 9 \right)} - 2 \ln{\left(\cos{\left(x \right)} \right)} 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{20}{4 x + 9} + \frac{24}{4 x + 4} - \frac{21}{3 x - 5} + \frac{7}{x - 9} 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{20}{4 x + 9} + \frac{24}{4 x + 4} - \frac{21}{3 x - 5} + \frac{7}{x - 9}\right)\left(\frac{\left(x - 9\right)^{7} \left(4 x + 4\right)^{6} e^{x}}{\left(3 x - 5\right)^{7} \left(4 x + 9\right)^{5} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{24}{4 x + 4} + \frac{7}{x - 9}2 \tan{\left(x \right)} - \frac{20}{4 x + 9} - \frac{21}{3 x - 5}\right)\left(\frac{\left(x - 9\right)^{7} \left(4 x + 4\right)^{6} e^{x}}{\left(3 x - 5\right)^{7} \left(4 x + 9\right)^{5} \cos^{2}{\left(x \right)}} \right)