Find the derivative of LaTeX: \displaystyle y = \frac{\left(8 - 2 x\right)^{5} e^{x} \sin^{2}{\left(x \right)}}{\left(4 - 6 x\right)^{4} \sqrt{8 x + 5} \cos^{8}{\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(8 - 2 x\right)^{5} e^{x} \sin^{2}{\left(x \right)}}{\left(4 - 6 x\right)^{4} \sqrt{8 x + 5} \cos^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 5 \ln{\left(8 - 2 x \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(4 - 6 x \right)} - \frac{\ln{\left(8 x + 5 \right)}}{2} - 8 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{8 x + 5} - \frac{10}{8 - 2 x} + \frac{24}{4 - 6 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{8 x + 5} - \frac{10}{8 - 2 x} + \frac{24}{4 - 6 x}\right)\left(\frac{\left(8 - 2 x\right)^{5} e^{x} \sin^{2}{\left(x \right)}}{\left(4 - 6 x\right)^{4} \sqrt{8 x + 5} \cos^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{2}{\tan{\left(x \right)}} - \frac{10}{8 - 2 x}8 \tan{\left(x \right)} - \frac{4}{8 x + 5} + \frac{24}{4 - 6 x}\right)\left(\frac{\left(8 - 2 x\right)^{5} e^{x} \sin^{2}{\left(x \right)}}{\left(4 - 6 x\right)^{4} \sqrt{8 x + 5} \cos^{8}{\left(x \right)}} \right)