Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 2\right)^{2} \left(4 x + 3\right)^{5} \left(8 x + 7\right)^{5} e^{x}}{\left(2 - 2 x\right)^{4} \sqrt{6 x + 5} \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(x - 2\right)^{2} \left(4 x + 3\right)^{5} \left(8 x + 7\right)^{5} e^{x}}{\left(2 - 2 x\right)^{4} \sqrt{6 x + 5} \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(x - 2 \right)} + 5 \ln{\left(4 x + 3 \right)} + 5 \ln{\left(8 x + 7 \right)}- 4 \ln{\left(2 - 2 x \right)} - \frac{\ln{\left(6 x + 5 \right)}}{2} - 6 \ln{\left(\cos{\left(x \right)} \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{40}{8 x + 7} - \frac{3}{6 x + 5} + \frac{20}{4 x + 3} + \frac{2}{x - 2} + \frac{8}{2 - 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{40}{8 x + 7} - \frac{3}{6 x + 5} + \frac{20}{4 x + 3} + \frac{2}{x - 2} + \frac{8}{2 - 2 x}\right)\left(\frac{\left(x - 2\right)^{2} \left(4 x + 3\right)^{5} \left(8 x + 7\right)^{5} e^{x}}{\left(2 - 2 x\right)^{4} \sqrt{6 x + 5} \cos^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{40}{8 x + 7} + \frac{20}{4 x + 3} + \frac{2}{x - 2}6 \tan{\left(x \right)} - \frac{3}{6 x + 5} + \frac{8}{2 - 2 x}\right)\left(\frac{\left(x - 2\right)^{2} \left(4 x + 3\right)^{5} \left(8 x + 7\right)^{5} e^{x}}{\left(2 - 2 x\right)^{4} \sqrt{6 x + 5} \cos^{6}{\left(x \right)}} \right)