Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 x + 7\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(9 - 4 x\right)^{5} \left(x + 9\right)^{4} \sqrt{\left(8 x + 3\right)^{7}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(3 x + 7\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(9 - 4 x\right)^{5} \left(x + 9\right)^{4} \sqrt{\left(8 x + 3\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(3 x + 7 \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 5 \ln{\left(9 - 4 x \right)} - 4 \ln{\left(x + 9 \right)} - \frac{7 \ln{\left(8 x + 3 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{28}{8 x + 3} + \frac{24}{3 x + 7} - \frac{4}{x + 9} + \frac{20}{9 - 4 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{28}{8 x + 3} + \frac{24}{3 x + 7} - \frac{4}{x + 9} + \frac{20}{9 - 4 x}\right)\left(\frac{\left(3 x + 7\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(9 - 4 x\right)^{5} \left(x + 9\right)^{4} \sqrt{\left(8 x + 3\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{24}{3 x + 7}-1 - \frac{28}{8 x + 3} - \frac{4}{x + 9} + \frac{20}{9 - 4 x}\right)\left(\frac{\left(3 x + 7\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(9 - 4 x\right)^{5} \left(x + 9\right)^{4} \sqrt{\left(8 x + 3\right)^{7}}} \right)