Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 3\right)^{5} \sqrt{\left(x + 4\right)^{3}} e^{- x} \sin^{7}{\left(x \right)}}{\left(- 4 x - 9\right)^{6} \left(2 x - 6\right)^{3}}

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 - 3\right)^{5} \sqrt{\left(x + 4\right)^{3}} e^{- x} \sin^{7}{\left(x \right)}}{\left(- 4 x - 9\right)^{6} \left(2 x - 6\right)^{3}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(x - 3 \right)} + \frac{3 \ln{\left(x + 4 \right)}}{2} + 7 \ln{\left(\sin{\left(x \right)} \right)}- x - 6 \ln{\left(- 4 x - 9 \right)} - 3 \ln{\left(2 x - 6 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{6}{2 x - 6} + \frac{3}{2 \left(x + 4\right)} + \frac{5}{x - 3} + \frac{24}{- 4 x - 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{6}{2 x - 6} + \frac{3}{2 \left(x + 4\right)} + \frac{5}{x - 3} + \frac{24}{- 4 x - 9}\right)\left(\frac{\left(x - 3\right)^{5} \sqrt{\left(x + 4\right)^{3}} e^{- x} \sin^{7}{\left(x \right)}}{\left(- 4 x - 9\right)^{6} \left(2 x - 6\right)^{3}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{\tan{\left(x \right)}} + \frac{3}{2 \left(x + 4\right)} + \frac{5}{x - 3}-1 - \frac{6}{2 x - 6} + \frac{24}{- 4 x - 9}\right)\left(\frac{\left(x - 3\right)^{5} \sqrt{\left(x + 4\right)^{3}} e^{- x} \sin^{7}{\left(x \right)}}{\left(- 4 x - 9\right)^{6} \left(2 x - 6\right)^{3}} \right)