Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 x - 9\right)^{8} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 7\right)^{4} \left(3 x + 3\right)^{4}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(4 x - 9\right)^{8} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 7\right)^{4} \left(3 x + 3\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 8 \ln{\left(4 x - 9 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 4 \ln{\left(x - 7 \right)} - 4 \ln{\left(3 x + 3 \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{32}{4 x - 9} - \frac{12}{3 x + 3} - \frac{4}{x - 7} 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{32}{4 x - 9} - \frac{12}{3 x + 3} - \frac{4}{x - 7}\right)\left(\frac{\left(4 x - 9\right)^{8} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 7\right)^{4} \left(3 x + 3\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 7 \tan{\left(x \right)} + 1 + \frac{32}{4 x - 9}- \frac{12}{3 x + 3} - \frac{4}{x - 7}\right)\left(\frac{\left(4 x - 9\right)^{8} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 7\right)^{4} \left(3 x + 3\right)^{4}} \right)