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

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(5 x + 6\right)^{8} e^{- x} \sin^{4}{\left(x \right)}}{\left(2 x - 8\right)^{6} \left(3 x + 3\right)^{8} \sqrt{\left(3 x + 4\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(5 x + 6 \right)} + 4 \ln{\left(\sin{\left(x \right)} \right)}- x - 6 \ln{\left(2 x - 8 \right)} - 8 \ln{\left(3 x + 3 \right)} - \frac{5 \ln{\left(3 x + 4 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{40}{5 x + 6} - \frac{15}{2 \left(3 x + 4\right)} - \frac{24}{3 x + 3} - \frac{12}{2 x - 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{40}{5 x + 6} - \frac{15}{2 \left(3 x + 4\right)} - \frac{24}{3 x + 3} - \frac{12}{2 x - 8}\right)\left(\frac{\left(5 x + 6\right)^{8} e^{- x} \sin^{4}{\left(x \right)}}{\left(2 x - 8\right)^{6} \left(3 x + 3\right)^{8} \sqrt{\left(3 x + 4\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{4}{\tan{\left(x \right)}} + \frac{40}{5 x + 6}-1 - \frac{15}{2 \left(3 x + 4\right)} - \frac{24}{3 x + 3} - \frac{12}{2 x - 8}\right)\left(\frac{\left(5 x + 6\right)^{8} e^{- x} \sin^{4}{\left(x \right)}}{\left(2 x - 8\right)^{6} \left(3 x + 3\right)^{8} \sqrt{\left(3 x + 4\right)^{5}}} \right)