Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 x - 3\right)^{8} \left(6 x - 5\right)^{2} e^{x}}{15625 x^{6} \left(x - 9\right)^{5} \sin^{6}{\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(5 x - 3\right)^{8} \left(6 x - 5\right)^{2} e^{x}}{15625 x^{6} \left(x - 9\right)^{5} \sin^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 8 \ln{\left(5 x - 3 \right)} + 2 \ln{\left(6 x - 5 \right)}- 6 \ln{\left(x \right)} - 5 \ln{\left(x - 9 \right)} - 6 \ln{\left(\sin{\left(x \right)} \right)} - 6 \ln{\left(5 \right)} 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{12}{6 x - 5} + \frac{40}{5 x - 3} - \frac{5}{x - 9} - \frac{6}{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{12}{6 x - 5} + \frac{40}{5 x - 3} - \frac{5}{x - 9} - \frac{6}{x}\right)\left(\frac{\left(5 x - 3\right)^{8} \left(6 x - 5\right)^{2} e^{x}}{15625 x^{6} \left(x - 9\right)^{5} \sin^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{12}{6 x - 5} + \frac{40}{5 x - 3}- \frac{6}{\tan{\left(x \right)}} - \frac{5}{x - 9} - \frac{6}{x}\right)\left(\frac{\left(5 x - 3\right)^{8} \left(6 x - 5\right)^{2} e^{x}}{15625 x^{6} \left(x - 9\right)^{5} \sin^{6}{\left(x \right)}} \right)