Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 5\right)^{5} \cos^{2}{\left(x \right)}}{\left(1 - 2 x\right)^{4} \sqrt{3 x + 8} \left(6 x - 2\right)^{2}}

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 - 5\right)^{5} \cos^{2}{\left(x \right)}}{\left(1 - 2 x\right)^{4} \sqrt{3 x + 8} \left(6 x - 2\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(x - 5 \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- 4 \ln{\left(1 - 2 x \right)} - \frac{\ln{\left(3 x + 8 \right)}}{2} - 2 \ln{\left(6 x - 2 \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)}} - \frac{12}{6 x - 2} - \frac{3}{2 \left(3 x + 8\right)} + \frac{5}{x - 5} + \frac{8}{1 - 2 x} 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)}} - \frac{12}{6 x - 2} - \frac{3}{2 \left(3 x + 8\right)} + \frac{5}{x - 5} + \frac{8}{1 - 2 x}\right)\left(\frac{\left(x - 5\right)^{5} \cos^{2}{\left(x \right)}}{\left(1 - 2 x\right)^{4} \sqrt{3 x + 8} \left(6 x - 2\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{5}{x - 5}- \frac{12}{6 x - 2} - \frac{3}{2 \left(3 x + 8\right)} + \frac{8}{1 - 2 x}\right)\left(\frac{\left(x - 5\right)^{5} \cos^{2}{\left(x \right)}}{\left(1 - 2 x\right)^{4} \sqrt{3 x + 8} \left(6 x - 2\right)^{2}} \right)