Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 - x\right)^{6} \left(- 5 x - 9\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(x - 3\right)^{8} \left(4 x - 1\right)^{8} \sqrt{\left(8 x + 1\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(4 - x\right)^{6} \left(- 5 x - 9\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(x - 3\right)^{8} \left(4 x - 1\right)^{8} \sqrt{\left(8 x + 1\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 6 \ln{\left(4 - x \right)} + 3 \ln{\left(- 5 x - 9 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(x - 3 \right)} - 8 \ln{\left(4 x - 1 \right)} - \frac{7 \ln{\left(8 x + 1 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{8 x + 1} - \frac{32}{4 x - 1} - \frac{8}{x - 3} - \frac{15}{- 5 x - 9} - \frac{6}{4 - x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{8 x + 1} - \frac{32}{4 x - 1} - \frac{8}{x - 3} - \frac{15}{- 5 x - 9} - \frac{6}{4 - x}\right)\left(\frac{\left(4 - x\right)^{6} \left(- 5 x - 9\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(x - 3\right)^{8} \left(4 x - 1\right)^{8} \sqrt{\left(8 x + 1\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{6}{\tan{\left(x \right)}} - \frac{15}{- 5 x - 9} - \frac{6}{4 - x}- \frac{28}{8 x + 1} - \frac{32}{4 x - 1} - \frac{8}{x - 3}\right)\left(\frac{\left(4 - x\right)^{6} \left(- 5 x - 9\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(x - 3\right)^{8} \left(4 x - 1\right)^{8} \sqrt{\left(8 x + 1\right)^{7}}} \right)