Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 - 5 x\right)^{7} \left(x - 6\right)^{7} e^{x} \sin^{6}{\left(x \right)}}{\left(x + 9\right)^{4} \sqrt{5 x + 4}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(3 - 5 x\right)^{7} \left(x - 6\right)^{7} e^{x} \sin^{6}{\left(x \right)}}{\left(x + 9\right)^{4} \sqrt{5 x + 4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 7 \ln{\left(3 - 5 x \right)} + 7 \ln{\left(x - 6 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(x + 9 \right)} - \frac{\ln{\left(5 x + 4 \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{5}{2 \left(5 x + 4\right)} - \frac{4}{x + 9} + \frac{7}{x - 6} - \frac{35}{3 - 5 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{5}{2 \left(5 x + 4\right)} - \frac{4}{x + 9} + \frac{7}{x - 6} - \frac{35}{3 - 5 x}\right)\left(\frac{\left(3 - 5 x\right)^{7} \left(x - 6\right)^{7} e^{x} \sin^{6}{\left(x \right)}}{\left(x + 9\right)^{4} \sqrt{5 x + 4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{6}{\tan{\left(x \right)}} + \frac{7}{x - 6} - \frac{35}{3 - 5 x}- \frac{5}{2 \left(5 x + 4\right)} - \frac{4}{x + 9}\right)\left(\frac{\left(3 - 5 x\right)^{7} \left(x - 6\right)^{7} e^{x} \sin^{6}{\left(x \right)}}{\left(x + 9\right)^{4} \sqrt{5 x + 4}} \right)