Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 9\right)^{4} \left(6 x - 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(4 x - 8\right)^{8} \left(6 x - 8\right)^{6} \sqrt{\left(9 x + 8\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(x + 9\right)^{4} \left(6 x - 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(4 x - 8\right)^{8} \left(6 x - 8\right)^{6} \sqrt{\left(9 x + 8\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 4 \ln{\left(x + 9 \right)} + 3 \ln{\left(6 x - 5 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(4 x - 8 \right)} - 6 \ln{\left(6 x - 8 \right)} - \frac{7 \ln{\left(9 x + 8 \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{63}{2 \left(9 x + 8\right)} + \frac{18}{6 x - 5} - \frac{36}{6 x - 8} - \frac{32}{4 x - 8} + \frac{4}{x + 9} 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{63}{2 \left(9 x + 8\right)} + \frac{18}{6 x - 5} - \frac{36}{6 x - 8} - \frac{32}{4 x - 8} + \frac{4}{x + 9}\right)\left(\frac{\left(x + 9\right)^{4} \left(6 x - 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(4 x - 8\right)^{8} \left(6 x - 8\right)^{6} \sqrt{\left(9 x + 8\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{6}{\tan{\left(x \right)}} + \frac{18}{6 x - 5} + \frac{4}{x + 9}- \frac{63}{2 \left(9 x + 8\right)} - \frac{36}{6 x - 8} - \frac{32}{4 x - 8}\right)\left(\frac{\left(x + 9\right)^{4} \left(6 x - 5\right)^{3} e^{x} \sin^{6}{\left(x \right)}}{\left(4 x - 8\right)^{8} \left(6 x - 8\right)^{6} \sqrt{\left(9 x + 8\right)^{7}}} \right)