Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 x - 5\right)^{8} \left(9 x - 8\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(x + 7\right)^{4} \sqrt{4 x + 4} \left(7 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(7 x - 5\right)^{8} \left(9 x - 8\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(x + 7\right)^{4} \sqrt{4 x + 4} \left(7 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(7 x - 5 \right)} + 8 \ln{\left(9 x - 8 \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 4 \ln{\left(x + 7 \right)} - \frac{\ln{\left(4 x + 4 \right)}}{2} - 7 \ln{\left(7 x + 3 \right)} 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{72}{9 x - 8} - \frac{49}{7 x + 3} + \frac{56}{7 x - 5} - \frac{2}{4 x + 4} - \frac{4}{x + 7} 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{72}{9 x - 8} - \frac{49}{7 x + 3} + \frac{56}{7 x - 5} - \frac{2}{4 x + 4} - \frac{4}{x + 7}\right)\left(\frac{\left(7 x - 5\right)^{8} \left(9 x - 8\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(x + 7\right)^{4} \sqrt{4 x + 4} \left(7 x + 3\right)^{7}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{72}{9 x - 8} + \frac{56}{7 x - 5}-1 - \frac{49}{7 x + 3} - \frac{2}{4 x + 4} - \frac{4}{x + 7}\right)\left(\frac{\left(7 x - 5\right)^{8} \left(9 x - 8\right)^{8} e^{- x} \cos^{5}{\left(x \right)}}{\left(x + 7\right)^{4} \sqrt{4 x + 4} \left(7 x + 3\right)^{7}} \right)