Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 x - 3\right)^{4} \left(8 x - 6\right)^{7} e^{- x}}{\left(x - 6\right)^{4} \left(3 x + 3\right)^{6} \sin^{4}{\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(7 x - 3\right)^{4} \left(8 x - 6\right)^{7} e^{- x}}{\left(x - 6\right)^{4} \left(3 x + 3\right)^{6} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(7 x - 3 \right)} + 7 \ln{\left(8 x - 6 \right)}- x - 4 \ln{\left(x - 6 \right)} - 6 \ln{\left(3 x + 3 \right)} - 4 \ln{\left(\sin{\left(x \right)} \right)} 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{56}{8 x - 6} + \frac{28}{7 x - 3} - \frac{18}{3 x + 3} - \frac{4}{x - 6} 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{56}{8 x - 6} + \frac{28}{7 x - 3} - \frac{18}{3 x + 3} - \frac{4}{x - 6}\right)\left(\frac{\left(7 x - 3\right)^{4} \left(8 x - 6\right)^{7} e^{- x}}{\left(x - 6\right)^{4} \left(3 x + 3\right)^{6} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{56}{8 x - 6} + \frac{28}{7 x - 3}-1 - \frac{4}{\tan{\left(x \right)}} - \frac{18}{3 x + 3} - \frac{4}{x - 6}\right)\left(\frac{\left(7 x - 3\right)^{4} \left(8 x - 6\right)^{7} e^{- x}}{\left(x - 6\right)^{4} \left(3 x + 3\right)^{6} \sin^{4}{\left(x \right)}} \right)